Quadratic Equations Sure Shot Questions for CBSE Class 10 Mathematics

Class 10 Mathematics Study Guide: CBSE Class 10 Quadratic Equations Sure Shot Questions Set 01

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Download Chapter 04 Quadratic Equations Study Material PDF

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PRACTICE QUESTIONS
CLASS X : CHAPTER - 4
QUADRATIC EQUATIONS
FACTORISATION METHOD

Solve the following quadratic equations:

1. x² + 11x + 30 = 0

Answer:
1. Here a × c = 1 × 30 = 30; two numbers with sum 11 and product 30 are 5 and 6.
2. Splitting the middle term: x² + 5x + 6x + 30 = 0.
3. Grouping: x(x + 5) + 6(x + 5) = 0.
4. (x + 6)(x + 5) = 0.
5. So x + 6 = 0 or x + 5 = 0, which gives x = -6 or x = -5.
6. Hence, the roots are -6 and -5.

Teacher's Note:
a) Write the equation in standard form ax² + bx + c = 0 before splitting the middle term.
b) Check the signs of the two numbers; their product has the sign of c.
c) Verify by substituting each root in the equation.

2. x² + 18x + 32 = 0

Answer:
1. Here a × c = 1 × 32 = 32; two numbers with sum 18 and product 32 are 2 and 16.
2. Splitting the middle term: x² + 2x + 16x + 32 = 0.
3. Grouping: x(x + 2) + 16(x + 2) = 0.
4. (x + 16)(x + 2) = 0.
5. So x + 16 = 0 or x + 2 = 0, which gives x = -16 or x = -2.
6. Hence, the roots are -16 and -2.

Teacher's Note:
a) After factorising, each bracket gives one root; do not forget to change the sign of the constant.
b) A quick check: sum of roots = -b/a and product = c/a.

3. x² + 7x - 18 = 0

Answer:
1. Here a × c = 1 × (-18) = -18; two numbers with sum 7 and product -18 are -2 and 9.
2. Splitting the middle term: x² - 2x + 9x - 18 = 0.
3. Grouping: x(x - 2) + 9(x - 2) = 0.
4. (x + 9)(x - 2) = 0.
5. So x + 9 = 0 or x - 2 = 0, which gives x = -9 or x = 2.
6. Hence, the roots are -9 and 2.

Teacher's Note:
a) Split the middle term so that the two parts add up to b and multiply to a × c.
b) Equate each factor to zero to get both roots.

4. x² + 5x - 6 = 0

Answer:
1. Here a × c = 1 × (-6) = -6; two numbers with sum 5 and product -6 are -1 and 6.
2. Splitting the middle term: x² - x + 6x - 6 = 0.
3. Grouping: x(x - 1) + 6(x - 1) = 0.
4. (x + 6)(x - 1) = 0.
5. So x + 6 = 0 or x - 1 = 0, which gives x = -6 or x = 1.
6. Hence, the roots are -6 and 1.

5. y² - 4y + 3 = 0

Answer:
1. Here a × c = 1 × 3 = 3; two numbers with sum -4 and product 3 are -1 and -3.
2. Splitting the middle term: y² - y - 3y + 3 = 0.
3. Grouping: y(y - 1) - 3(y - 1) = 0.
4. (y - 3)(y - 1) = 0.
5. So y - 3 = 0 or y - 1 = 0, which gives y = 3 or y = 1.
6. Hence, the roots are 3 and 1.

Teacher's Note:
a) After factorising, each bracket gives one root; do not forget to change the sign of the constant.
b) A quick check: sum of roots = -b/a and product = c/a.

6. x² - 21x + 108 = 0

Answer:
1. Here a × c = 1 × 108 = 108; two numbers with sum -21 and product 108 are -9 and -12.
2. Splitting the middle term: x² - 9x - 12x + 108 = 0.
3. Grouping: x(x - 9) - 12(x - 9) = 0.
4. (x - 12)(x - 9) = 0.
5. So x - 12 = 0 or x - 9 = 0, which gives x = 12 or x = 9.
6. Hence, the roots are 12 and 9.

Teacher's Note:
a) Split the middle term so that the two parts add up to b and multiply to a × c.
b) Equate each factor to zero to get both roots.

7. x² - 11x - 80 = 0

Answer:
1. Here a × c = 1 × (-80) = -80; two numbers with sum -11 and product -80 are 5 and -16.
2. Splitting the middle term: x² + 5x - 16x - 80 = 0.
3. Grouping: x(x + 5) - 16(x + 5) = 0.
4. (x - 16)(x + 5) = 0.
5. So x - 16 = 0 or x + 5 = 0, which gives x = 16 or x = -5.
6. Hence, the roots are 16 and -5.

8. x² - x - 156 = 0

Answer:
1. Here a × c = 1 × (-156) = -156; two numbers with sum -1 and product -156 are 12 and -13.
2. Splitting the middle term: x² + 12x - 13x - 156 = 0.
3. Grouping: x(x + 12) - 13(x + 12) = 0.
4. (x - 13)(x + 12) = 0.
5. So x - 13 = 0 or x + 12 = 0, which gives x = 13 or x = -12.
6. Hence, the roots are 13 and -12.

Teacher's Note:
a) After factorising, each bracket gives one root; do not forget to change the sign of the constant.
b) A quick check: sum of roots = -b/a and product = c/a.

9. z² - 32z - 105 = 0

Answer:
1. Here a × c = 1 × (-105) = -105; two numbers with sum -32 and product -105 are 3 and -35.
2. Splitting the middle term: z² + 3z - 35z - 105 = 0.
3. Grouping: z(z + 3) - 35(z + 3) = 0.
4. (z - 35)(z + 3) = 0.
5. So z - 35 = 0 or z + 3 = 0, which gives z = 35 or z = -3.
6. Hence, the roots are 35 and -3.

Teacher's Note:
a) Split the middle term so that the two parts add up to b and multiply to a × c.
b) Equate each factor to zero to get both roots.

10. 40 + 3x - x² = 0

Answer:
1. Rearranging and multiplying by -1: x² - 3x - 40 = 0.
2. Here a × c = 1 × (-40) = -40; two numbers with sum -3 and product -40 are 5 and -8.
3. Splitting the middle term: x² + 5x - 8x - 40 = 0.
4. Grouping: x(x + 5) - 8(x + 5) = 0.
5. (x - 8)(x + 5) = 0.
6. So x - 8 = 0 or x + 5 = 0, which gives x = 8 or x = -5.
7. Hence, the roots are 8 and -5.

11. 6 - x - x² = 0

Answer:
1. Rearranging and multiplying by -1: x² + x - 6 = 0.
2. Here a × c = 1 × (-6) = -6; two numbers with sum 1 and product -6 are -2 and 3.
3. Splitting the middle term: x² - 2x + 3x - 6 = 0.
4. Grouping: x(x - 2) + 3(x - 2) = 0.
5. (x + 3)(x - 2) = 0.
6. So x + 3 = 0 or x - 2 = 0, which gives x = -3 or x = 2.
7. Hence, the roots are -3 and 2.

Teacher's Note:
a) After factorising, each bracket gives one root; do not forget to change the sign of the constant.
b) A quick check: sum of roots = -b/a and product = c/a.

12. 7x² + 49x + 84 = 0

Answer:
1. Here a × c = 7 × 84 = 588; two numbers with sum 49 and product 588 are 21 and 28.
2. Splitting the middle term: 7x² + 21x + 28x + 84 = 0.
3. Grouping: 7x(x + 3) + 28(x + 3) = 0.
4. (7x + 28)(x + 3) = 0.
5. So 7x + 28 = 0 or x + 3 = 0, which gives x = -4 or x = -3.
6. Hence, the roots are -4 and -3.

Teacher's Note:
a) Split the middle term so that the two parts add up to b and multiply to a × c.
b) Equate each factor to zero to get both roots.

13. m² + 17mn - 84n² = 0

Answer:
1. Here a × c = -84n² and we need two terms with sum 17n and product -84n²; they are 21n and -4n.
2. Splitting the middle term: m² + 21mn - 4mn - 84n² = 0.
3. Grouping: m(m + 21n) - 4n(m + 21n) = 0.
4. (m - 4n)(m + 21n) = 0.
5. So m - 4n = 0 or m + 21n = 0, which gives m = 4n or m = -21n.
6. Hence, m = 4n or m = -21n.

Teacher's Note:
a) Split the middle term using expressions in a and b whose sum and product match.
b) Final roots are expressions in a and b; leave them in simplest form.
c) Verify using sum of roots = -b/a.

14. 5x² + 16x + 3 = 0

Answer:
1. Here a × c = 5 × 3 = 15; two numbers with sum 16 and product 15 are 1 and 15.
2. Splitting the middle term: 5x² + x + 15x + 3 = 0.
3. Grouping: x(5x + 1) + 3(5x + 1) = 0.
4. (x + 3)(5x + 1) = 0.
5. So x + 3 = 0 or 5x + 1 = 0, which gives x = -3 or x = \( - \frac{1}{5} \).
6. Hence, the roots are -3 and \( - \frac{1}{5} \).

15. 6x² + 17x +12 = 0

Answer:
1. Here a × c = 6 × 12 = 72; two numbers with sum 17 and product 72 are 8 and 9.
2. Splitting the middle term: 6x² + 8x + 9x + 12 = 0.
3. Grouping: 2x(3x + 4) + 3(3x + 4) = 0.
4. (2x + 3)(3x + 4) = 0.
5. So 2x + 3 = 0 or 3x + 4 = 0, which gives x = \( - \frac{3}{2} \) or x = \( - \frac{4}{3} \).
6. Hence, the roots are \( - \frac{3}{2} \) and \( - \frac{4}{3} \).

Teacher's Note:
a) After factorising, each bracket gives one root; do not forget to change the sign of the constant.
b) A quick check: sum of roots = -b/a and product = c/a.

16. 9x² + 18x + 8 = 0

Answer:
1. Here a × c = 9 × 8 = 72; two numbers with sum 18 and product 72 are 6 and 12.
2. Splitting the middle term: 9x² + 6x + 12x + 8 = 0.
3. Grouping: 3x(3x + 2) + 4(3x + 2) = 0.
4. (3x + 4)(3x + 2) = 0.
5. So 3x + 4 = 0 or 3x + 2 = 0, which gives x = \( - \frac{4}{3} \) or x = \( - \frac{2}{3} \).
6. Hence, the roots are \( - \frac{4}{3} \) and \( - \frac{2}{3} \).

Teacher's Note:
a) Split the middle term so that the two parts add up to b and multiply to a × c.
b) Equate each factor to zero to get both roots.

17. 14x² + 9x + 1 = 0

Answer:
1. Here a × c = 14 × 1 = 14; two numbers with sum 9 and product 14 are 2 and 7.
2. Splitting the middle term: 14x² + 2x + 7x + 1 = 0.
3. Grouping: 2x(7x + 1) + (7x + 1) = 0.
4. (2x + 1)(7x + 1) = 0.
5. So 2x + 1 = 0 or 7x + 1 = 0, which gives x = \( - \frac{1}{2} \) or x = \( - \frac{1}{7} \).
6. Hence, the roots are \( - \frac{1}{2} \) and \( - \frac{1}{7} \).

18. 2x² + 3x - 90 = 0

Answer:
1. Here a × c = 2 × (-90) = -180; two numbers with sum 3 and product -180 are -12 and 15.
2. Splitting the middle term: 2x² - 12x + 15x - 90 = 0.
3. Grouping: 2x(x - 6) + 15(x - 6) = 0.
4. (2x + 15)(x - 6) = 0.
5. So 2x + 15 = 0 or x - 6 = 0, which gives x = \( - \frac{15}{2} \) or x = 6.
6. Hence, the roots are \( - \frac{15}{2} \) and 6.

Teacher's Note:
a) After factorising, each bracket gives one root; do not forget to change the sign of the constant.
b) A quick check: sum of roots = -b/a and product = c/a.

19. 2x² + 11x - 21 = 0

Answer:
1. Here a × c = 2 × (-21) = -42; two numbers with sum 11 and product -42 are -3 and 14.
2. Splitting the middle term: 2x² - 3x + 14x - 21 = 0.
3. Grouping: x(2x - 3) + 7(2x - 3) = 0.
4. (x + 7)(2x - 3) = 0.
5. So x + 7 = 0 or 2x - 3 = 0, which gives x = -7 or x = \( \frac{3}{2} \).
6. Hence, the roots are -7 and \( \frac{3}{2} \).

Teacher's Note:
a) Split the middle term so that the two parts add up to b and multiply to a × c.
b) Equate each factor to zero to get both roots.

20. 3x² - 14x + 8 = 0

Answer:
1. Here a × c = 3 × 8 = 24; two numbers with sum -14 and product 24 are -2 and -12.
2. Splitting the middle term: 3x² - 2x - 12x + 8 = 0.
3. Grouping: x(3x - 2) - 4(3x - 2) = 0.
4. (x - 4)(3x - 2) = 0.
5. So x - 4 = 0 or 3x - 2 = 0, which gives x = 4 or x = \( \frac{2}{3} \).
6. Hence, the roots are 4 and \( \frac{2}{3} \).

21. 18x² + 3x - 10 = 0

Answer:
1. Here a × c = 18 × (-10) = -180; two numbers with sum 3 and product -180 are -12 and 15.
2. Splitting the middle term: 18x² - 12x + 15x - 10 = 0.
3. Grouping: 6x(3x - 2) + 5(3x - 2) = 0.
4. (6x + 5)(3x - 2) = 0.
5. So 6x + 5 = 0 or 3x - 2 = 0, which gives x = \( - \frac{5}{6} \) or x = \( \frac{2}{3} \).
6. Hence, the roots are \( - \frac{5}{6} \) and \( \frac{2}{3} \).

Teacher's Note:
a) After factorising, each bracket gives one root; do not forget to change the sign of the constant.
b) A quick check: sum of roots = -b/a and product = c/a.

22. 15x² + 2x - 8 = 0

Answer:
1. Here a × c = 15 × (-8) = -120; two numbers with sum 2 and product -120 are -10 and 12.
2. Splitting the middle term: 15x² - 10x + 12x - 8 = 0.
3. Grouping: 5x(3x - 2) + 4(3x - 2) = 0.
4. (5x + 4)(3x - 2) = 0.
5. So 5x + 4 = 0 or 3x - 2 = 0, which gives x = \( - \frac{4}{5} \) or x = \( \frac{2}{3} \).
6. Hence, the roots are \( - \frac{4}{5} \) and \( \frac{2}{3} \).

Teacher's Note:
a) Split the middle term so that the two parts add up to b and multiply to a × c.
b) Equate each factor to zero to get both roots.

23. 6x² + 11x - 10 = 0

Answer:
1. Here a × c = 6 × (-10) = -60; two numbers with sum 11 and product -60 are -4 and 15.
2. Splitting the middle term: 6x² - 4x + 15x - 10 = 0.
3. Grouping: 2x(3x - 2) + 5(3x - 2) = 0.
4. (2x + 5)(3x - 2) = 0.
5. So 2x + 5 = 0 or 3x - 2 = 0, which gives x = \( - \frac{5}{2} \) or x = \( \frac{2}{3} \).
6. Hence, the roots are \( - \frac{5}{2} \) and \( \frac{2}{3} \).

24. 30x² + 7x - 15 = 0

Answer:
1. Here a × c = 30 × (-15) = -450; two numbers with sum 7 and product -450 are -18 and 25.
2. Splitting the middle term: 30x² - 18x + 25x - 15 = 0.
3. Grouping: 6x(5x - 3) + 5(5x - 3) = 0.
4. (6x + 5)(5x - 3) = 0.
5. So 6x + 5 = 0 or 5x - 3 = 0, which gives x = \( - \frac{5}{6} \) or x = \( \frac{3}{5} \).
6. Hence, the roots are \( - \frac{5}{6} \) and \( \frac{3}{5} \).

Teacher's Note:
a) After factorising, each bracket gives one root; do not forget to change the sign of the constant.
b) A quick check: sum of roots = -b/a and product = c/a.

25. 24x² - 41x + 12 = 0

Answer:
1. Here a × c = 24 × 12 = 288; two numbers with sum -41 and product 288 are -9 and -32.
2. Splitting the middle term: 24x² - 9x - 32x + 12 = 0.
3. Grouping: 3x(8x - 3) - 4(8x - 3) = 0.
4. (3x - 4)(8x - 3) = 0.
5. So 3x - 4 = 0 or 8x - 3 = 0, which gives x = \( \frac{4}{3} \) or x = \( \frac{3}{8} \).
6. Hence, the roots are \( \frac{4}{3} \) and \( \frac{3}{8} \).

Teacher's Note:
a) Split the middle term so that the two parts add up to b and multiply to a × c.
b) Equate each factor to zero to get both roots.

26. 2x² - 7x - 15 = 0

Answer:
1. Here a × c = 2 × (-15) = -30; two numbers with sum -7 and product -30 are 3 and -10.
2. Splitting the middle term: 2x² + 3x - 10x - 15 = 0.
3. Grouping: x(2x + 3) - 5(2x + 3) = 0.
4. (x - 5)(2x + 3) = 0.
5. So x - 5 = 0 or 2x + 3 = 0, which gives x = 5 or x = \( - \frac{3}{2} \).
6. Hence, the roots are 5 and \( - \frac{3}{2} \).

27. 6x² + 11x - 10 = 0

Teacher's Note:
a) After factorising, each bracket gives one root; do not forget to change the sign of the constant.
b) A quick check: sum of roots = -b/a and product = c/a.

28. 10x² - 9x - 7 = 0

Answer:
1. Here a × c = 10 × (-7) = -70; two numbers with sum -9 and product -70 are 5 and -14.
2. Splitting the middle term: 10x² + 5x - 14x - 7 = 0.
3. Grouping: 5x(2x + 1) - 7(2x + 1) = 0.
4. (5x - 7)(2x + 1) = 0.
5. So 5x - 7 = 0 or 2x + 1 = 0, which gives x = \( \frac{7}{5} \) or x = \( - \frac{1}{2} \).
6. Hence, the roots are \( \frac{7}{5} \) and \( - \frac{1}{2} \).

Teacher's Note:
a) Split the middle term so that the two parts add up to b and multiply to a × c.
b) Equate each factor to zero to get both roots.

29. 5x² - 16x - 21 = 0

Answer:
1. Here a × c = 5 × (-21) = -105; two numbers with sum -16 and product -105 are 5 and -21.
2. Splitting the middle term: 5x² + 5x - 21x - 21 = 0.
3. Grouping: 5x(x + 1) - 21(x + 1) = 0.
4. (5x - 21)(x + 1) = 0.
5. So 5x - 21 = 0 or x + 1 = 0, which gives x = \( \frac{21}{5} \) or x = -1.
6. Hence, the roots are \( \frac{21}{5} \) and -1.

30. 2x² - x - 21 = 0

Answer:
1. Here a × c = 2 × (-21) = -42; two numbers with sum -1 and product -42 are 6 and -7.
2. Splitting the middle term: 2x² + 6x - 7x - 21 = 0.
3. Grouping: 2x(x + 3) - 7(x + 3) = 0.
4. (2x - 7)(x + 3) = 0.
5. So 2x - 7 = 0 or x + 3 = 0, which gives x = \( \frac{7}{2} \) or x = -3.
6. Hence, the roots are \( \frac{7}{2} \) and -3.

Teacher's Note:
a) After factorising, each bracket gives one root; do not forget to change the sign of the constant.
b) A quick check: sum of roots = -b/a and product = c/a.

31. 15x² - x - 28 = 0

Answer:
1. Here a × c = 15 × (-28) = -420; two numbers with sum -1 and product -420 are 20 and -21.
2. Splitting the middle term: 15x² + 20x - 21x - 28 = 0.
3. Grouping: 5x(3x + 4) - 7(3x + 4) = 0.
4. (5x - 7)(3x + 4) = 0.
5. So 5x - 7 = 0 or 3x + 4 = 0, which gives x = \( \frac{7}{5} \) or x = \( - \frac{4}{3} \).
6. Hence, the roots are \( \frac{7}{5} \) and \( - \frac{4}{3} \).

Teacher's Note:
a) Split the middle term so that the two parts add up to b and multiply to a × c.
b) Equate each factor to zero to get both roots.

32. 8a² - 27ab + 9b² = 0

Answer:
1. Here a × c = 72b² and we need two terms with sum -27b and product 72b²; they are -24b and -3b.
2. Splitting the middle term: 8a² - 24ab - 3ab + 9b² = 0.
3. Grouping: 8a(a - 3b) - 3b(a - 3b) = 0.
4. (8a - 3b)(a - 3b) = 0.
5. So 8a - 3b = 0 or a - 3b = 0, which gives a = \( \frac{3b}{8} \) or a = 3b.
6. Hence, a = \( \frac{3b}{8} \) or a = 3b.

Teacher's Note:
a) Treat a and b as constants and factorise exactly as with numbers.
b) Check by substituting a root back in the equation.

33. 5x² + 33xy - 14y² = 0

Answer:
1. Here a × c = -70y² and we need two terms with sum 33y and product -70y²; they are 35y and -2y.
2. Splitting the middle term: 5x² + 35xy - 2xy - 14y² = 0.
3. Grouping: 5x(x + 7y) - 2y(x + 7y) = 0.
4. (5x - 2y)(x + 7y) = 0.
5. So 5x - 2y = 0 or x + 7y = 0, which gives x = \( \frac{2y}{5} \) or x = -7y.
6. Hence, x = \( \frac{2y}{5} \) or x = -7y.

34. 3x³ - x² - 10x = 0

Answer:
1. Taking x common: x(3x² - x - 10) = 0, so x = 0 or 3x² - x - 10 = 0.
2. Here a × c = 3 × (-10) = -30; two numbers with sum -1 and product -30 are 5 and -6.
3. Splitting the middle term: 3x² + 5x - 6x - 10 = 0.
4. Grouping: x(3x + 5) - 2(3x + 5) = 0.
5. (x - 2)(3x + 5) = 0.
6. So x - 2 = 0 or 3x + 5 = 0, which gives x = 2 or x = \( - \frac{5}{3} \).
7. Hence, the roots are 0, 2 and \( - \frac{5}{3} \).

35. x² + 9x + 18 = 0

Answer:
1. Here a × c = 1 × 18 = 18; two numbers with sum 9 and product 18 are 3 and 6.
2. Splitting the middle term: x² + 3x + 6x + 18 = 0.
3. Grouping: x(x + 3) + 6(x + 3) = 0.
4. (x + 6)(x + 3) = 0.
5. So x + 6 = 0 or x + 3 = 0, which gives x = -6 or x = -3.
6. Hence, the roots are -6 and -3.

Teacher's Note:
a) After factorising, each bracket gives one root; do not forget to change the sign of the constant.
b) A quick check: sum of roots = -b/a and product = c/a.

36. x² + 5x - 24 = 0

Answer:
1. Here a × c = 1 × (-24) = -24; two numbers with sum 5 and product -24 are -3 and 8.
2. Splitting the middle term: x² - 3x + 8x - 24 = 0.
3. Grouping: x(x - 3) + 8(x - 3) = 0.
4. (x + 8)(x - 3) = 0.
5. So x + 8 = 0 or x - 3 = 0, which gives x = -8 or x = 3.
6. Hence, the roots are -8 and 3.

Teacher's Note:
a) Split the middle term so that the two parts add up to b and multiply to a × c.
b) Equate each factor to zero to get both roots.

37. x² - 4x - 21 = 0

Answer:
1. Here a × c = 1 × (-21) = -21; two numbers with sum -4 and product -21 are 3 and -7.
2. Splitting the middle term: x² + 3x - 7x - 21 = 0.
3. Grouping: x(x + 3) - 7(x + 3) = 0.
4. (x - 7)(x + 3) = 0.
5. So x - 7 = 0 or x + 3 = 0, which gives x = 7 or x = -3.
6. Hence, the roots are 7 and -3.

38. 6x² + 7x - 3 = 0

Answer:
1. Here a × c = 6 × (-3) = -18; two numbers with sum 7 and product -18 are -2 and 9.
2. Splitting the middle term: 6x² - 2x + 9x - 3 = 0.
3. Grouping: 2x(3x - 1) + 3(3x - 1) = 0.
4. (2x + 3)(3x - 1) = 0.
5. So 2x + 3 = 0 or 3x - 1 = 0, which gives x = \( - \frac{3}{2} \) or x = \( \frac{1}{3} \).
6. Hence, the roots are \( - \frac{3}{2} \) and \( \frac{1}{3} \).

Teacher's Note:
a) After factorising, each bracket gives one root; do not forget to change the sign of the constant.
b) A quick check: sum of roots = -b/a and product = c/a.

39. 2x² - 7x - 39 = 0

Answer:
1. Here a × c = 2 × (-39) = -78; two numbers with sum -7 and product -78 are 6 and -13.
2. Splitting the middle term: 2x² + 6x - 13x - 39 = 0.
3. Grouping: 2x(x + 3) - 13(x + 3) = 0.
4. (2x - 13)(x + 3) = 0.
5. So 2x - 13 = 0 or x + 3 = 0, which gives x = \( \frac{13}{2} \) or x = -3.
6. Hence, the roots are \( \frac{13}{2} \) and -3.

Teacher's Note:
a) Split the middle term so that the two parts add up to b and multiply to a × c.
b) Equate each factor to zero to get both roots.

40. 9x² - 22x + 8 = 0

Answer:
1. Here a × c = 9 × 8 = 72; two numbers with sum -22 and product 72 are -4 and -18.
2. Splitting the middle term: 9x² - 4x - 18x + 8 = 0.
3. Grouping: x(9x - 4) - 2(9x - 4) = 0.
4. (x - 2)(9x - 4) = 0.
5. So x - 2 = 0 or 9x - 4 = 0, which gives x = 2 or x = \( \frac{4}{9} \).
6. Hence, the roots are 2 and \( \frac{4}{9} \).

41. 6x² + 40 = 31x

Answer:
1. Writing in standard form: 6x² - 31x + 40 = 0.
2. Here a × c = 6 × 40 = 240; two numbers with sum -31 and product 240 are -15 and -16.
3. Splitting the middle term: 6x² - 15x - 16x + 40 = 0.
4. Grouping: 3x(2x - 5) - 8(2x - 5) = 0.
5. (3x - 8)(2x - 5) = 0.
6. So 3x - 8 = 0 or 2x - 5 = 0, which gives x = \( \frac{8}{3} \) or x = \( \frac{5}{2} \).
7. Hence, the roots are \( \frac{8}{3} \) and \( \frac{5}{2} \).

Teacher's Note:
a) After factorising, each bracket gives one root; do not forget to change the sign of the constant.
b) A quick check: sum of roots = -b/a and product = c/a.

42. 36x² - 12ax + (a² - b²) = 0

Answer:
1. Since 36x² - 12ax + a² = (6x - a)², the equation is (6x - a)² - b² = 0.
2. Using p² - q² = (p - q)(p + q): (6x - a - b)(6x - a + b) = 0.
3. So 6x - a - b = 0 or 6x - a + b = 0.
4. x = \( \frac{a + b}{6} \) or x = \( \frac{a - b}{6} \).
5. Hence, the roots are \( \frac{a + b}{6} \) and \( \frac{a - b}{6} \).

Teacher's Note:
a) Treat a and b as constants and factorise exactly as with numbers.
b) Check by substituting a root back in the equation.

43. 8x² - 22x - 21 = 0

Answer:
1. Here a × c = 8 × (-21) = -168; two numbers with sum -22 and product -168 are 6 and -28.
2. Splitting the middle term: 8x² + 6x - 28x - 21 = 0.
3. Grouping: 2x(4x + 3) - 7(4x + 3) = 0.
4. (2x - 7)(4x + 3) = 0.
5. So 2x - 7 = 0 or 4x + 3 = 0, which gives x = \( \frac{7}{2} \) or x = \( - \frac{3}{4} \).
6. Hence, the roots are \( \frac{7}{2} \) and \( - \frac{3}{4} \).

Teacher's Note:
a) Split the middle term so that the two parts add up to b and multiply to a × c.
b) Equate each factor to zero to get both roots.

44. 2x² - x + \( \frac{1}{8} \) = 0

Answer:
1. Multiplying by 8: 16x² - 8x + 1 = 0.
2. Here a × c = 16 × 1 = 16; two numbers with sum -8 and product 16 are -4 and -4.
3. Splitting the middle term: 16x² - 4x - 4x + 1 = 0.
4. Grouping: 4x(4x - 1) - (4x - 1) = 0.
5. (4x - 1)(4x - 1) = 0.
6. Hence, (4x - 1)² = 0 and the roots are equal: x = \( \frac{1}{4} \).
7. Hence, the roots are \( \frac{1}{4} \) (equal roots).

45. 4√3x² + 5x - 2√3 = 0

CBSE Class 10 Quadratic Equations Sure Shot Questions Set A

Answer:
1. Here a × c = 4√3 × (-2√3) = -24; two numbers with sum 5 and product -24 are 8 and -3.
2. Splitting the middle term: 4√3x² + 8x - 3x - 2√3 = 0.
3. Grouping: 4x(√3x + 2) - √3(√3x + 2) = 0.
4. (4x - √3)(√3x + 2) = 0.
5. So x = \( \frac{\sqrt{3}}{4} \) or x = \( -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3} \).
6. Hence, the roots are \( \frac{\sqrt{3}}{4} \) and \( -\frac{2\sqrt{3}}{3} \).

Teacher's Note:
a) Find two numbers whose sum is the middle coefficient and whose product is a × c, even when a and c contain radicals.
b) Group the terms so that both groups share the same bracket.
c) Rationalise answers such as 2/√3 to 2√3/3.

Solve the following by Factorisation method:

1. √2x² + 7x + 5√2 = 0

Answer:
1. 1. Here a × c = √2 × 5√2 = 10; two numbers with sum 7 and product 10 are 2 and 5.
2. 2. Splitting the middle term: √2x² + 2x + 5x + 5√2 = 0.
3. 3. Grouping: √2x(x + √2) + 5(x + √2) = 0.
4. 4. (√2x + 5)(x + √2) = 0.
5. 5. So x = \( -\frac{5}{\sqrt{2}} = -\frac{5\sqrt{2}}{2} \) or x = -√2.
6. 6. Hence, the roots are \( -\frac{5\sqrt{2}}{2} \) and -√2.

Teacher's Note:
a) Square roots such as √3 are treated like numbers when splitting the middle term.
b) Rationalise the denominator in the final roots.

2. 2x - \( \frac{3}{x} \) = 1

Answer:
1. Multiplying both sides by the LCD, x, to remove the denominators and simplifying: 2x² - x - 3 = 0.
2. Here a × c = 2 × (-3) = -6; two numbers with sum -1 and product -6 are 2 and -3.
3. Splitting the middle term: 2x² + 2x - 3x - 3 = 0.
4. Grouping: 2x(x + 1) - 3(x + 1) = 0.
5. (2x - 3)(x + 1) = 0.
6. So 2x - 3 = 0 or x + 1 = 0, which gives x = \( \frac{3}{2} \) or x = -1.
7. Neither value makes a denominator zero, so both are valid. Hence, the roots are \( \frac{3}{2} \) and -1.

Teacher's Note:
a) Cross-multiplying carefully avoids sign errors; expand every bracket before collecting terms.
b) A root that makes a denominator zero must be rejected.
c) Verify the final roots in the original equation.

3. \( \frac{4}{x} \) - 3 = \( \frac{5}{2x + 3} \), x ≠ 0, \( \frac{-3}{2} \)

Answer:
1. Multiplying both sides by the LCD, x(2x + 3), to remove the denominators and simplifying: x² + x - 2 = 0.
2. Here a × c = 1 × (-2) = -2; two numbers with sum 1 and product -2 are -1 and 2.
3. Splitting the middle term: x² - x + 2x - 2 = 0.
4. Grouping: x(x - 1) + 2(x - 1) = 0.
5. (x + 2)(x - 1) = 0.
6. So x + 2 = 0 or x - 1 = 0, which gives x = -2 or x = 1.
7. Neither value makes a denominator zero, so both are valid. Hence, the roots are -2 and 1.

Teacher's Note:
a) Reduce the equation to ax² + bx + c = 0 and divide by any common factor first.
b) Both roots are valid only if neither makes a denominator zero.

4. \( \frac{x}{x + 1} \) + \( \frac{x + 1}{x} \) = \( \frac{34}{15} \), x ≠ -1 and x ≠ 0

Answer:
1. Multiplying both sides by the LCD, 15x(x + 1), to remove the denominators and simplifying: 4x² + 4x - 15 = 0.
2. Here a × c = 4 × (-15) = -60; two numbers with sum 4 and product -60 are -6 and 10.
3. Splitting the middle term: 4x² - 6x + 10x - 15 = 0.
4. Grouping: 2x(2x - 3) + 5(2x - 3) = 0.
5. (2x + 5)(2x - 3) = 0.
6. So 2x + 5 = 0 or 2x - 3 = 0, which gives x = \( - \frac{5}{2} \) or x = \( \frac{3}{2} \).
7. Neither value makes a denominator zero, so both are valid. Hence, the roots are \( - \frac{5}{2} \) and \( \frac{3}{2} \).

Teacher's Note:
a) Multiply by the LCD to remove the denominators, then simplify to a quadratic.
b) Always compare the roots with the values excluded in the question.

5. \( \frac{x + 3}{x + 2} \) = \( \frac{3x - 7}{2x - 3} \)

Answer:
1. Cross-multiplying: (x + 3)(2x - 3) = (3x - 7)(x + 2), so 2x² + 3x - 9 = 3x² - x - 14, which gives x² - 4x - 5 = 0.
2. Here a × c = 1 × (-5) = -5; two numbers with sum -4 and product -5 are 1 and -5.
3. Splitting the middle term: x² + x - 5x - 5 = 0.
4. Grouping: x(x + 1) - 5(x + 1) = 0.
5. (x - 5)(x + 1) = 0.
6. So x - 5 = 0 or x + 1 = 0, which gives x = 5 or x = -1.
7. Neither value makes a denominator zero, so both are valid. Hence, the roots are 5 and -1.

6. \( \frac{x - 1}{x - 2} \) + \( \frac{x - 3}{x - 4} \) = 3\( \frac{1}{3} \) (x ≠ 2, 4)

Answer:
1. Writing 3\( \frac{1}{3} \) = \( \frac{10}{3} \) and multiplying by 3(x - 2)(x - 4): 3(x - 1)(x - 4) + 3(x - 3)(x - 2) = 10(x - 2)(x - 4), which simplifies to 2x² - 15x + 25 = 0.
2. Here a × c = 2 × 25 = 50; two numbers with sum -15 and product 50 are -5 and -10.
3. Splitting the middle term: 2x² - 5x - 10x + 25 = 0.
4. Grouping: x(2x - 5) - 5(2x - 5) = 0.
5. (x - 5)(2x - 5) = 0.
6. So x - 5 = 0 or 2x - 5 = 0, which gives x = 5 or x = \( \frac{5}{2} \).
7. Neither value makes a denominator zero, so both are valid. Hence, the roots are 5 and \( \frac{5}{2} \).

Teacher's Note:
a) Reduce the equation to ax² + bx + c = 0 and divide by any common factor first.
b) Both roots are valid only if neither makes a denominator zero.

7. \( \frac{1}{a + b + x} \) = \( \frac{1}{a} \) + \( \frac{1}{b} \) + \( \frac{1}{x} \), [x ≠ 0, -(a + b)]

Answer:
1. 1. Moving \( \frac{1}{x} \) to the left: \( \frac{1}{a + b + x} \) - \( \frac{1}{x} \) = \( \frac{1}{a} \) + \( \frac{1}{b} \).
2. 2. Taking the LCM on both sides: \( \frac{x - (a + b + x)}{x(a + b + x)} = \frac{a + b}{ab} \), so \( \frac{-(a + b)}{x(a + b + x)} = \frac{a + b}{ab} \).
3. 3. Dividing both sides by (a + b) and cross-multiplying: -ab = x(a + b + x).
4. 4. x² + (a + b)x + ab = 0.
5. 5. Splitting the middle term: x² + ax + bx + ab = 0, so x(x + a) + b(x + a) = 0.
6. 6. (x + a)(x + b) = 0.
7. 7. Hence, x = -a or x = -b.

8. 2\( \left(\frac{2x - 1}{x + 3}\right) \) - 3\( \left(\frac{x + 3}{2x - 1}\right) \) = 5, x ≠ -3, \( \frac{1}{2} \)

Answer:
1. 1. Put y = \( \frac{2x - 1}{x + 3} \). Then \( \frac{x + 3}{2x - 1} \) = \( \frac{1}{y} \) and the equation becomes 2y - \( \frac{3}{y} \) = 5.
2. 2. Multiplying by y: 2y² - 5y - 3 = 0.
3. 3. Splitting the middle term: 2y² - 6y + y - 3 = 0, so 2y(y - 3) + 1(y - 3) = 0.
4. 4. (2y + 1)(y - 3) = 0, so y = 3 or y = \( \frac{-1}{2} \).
5. 5. If y = 3: 2x - 1 = 3(x + 3), so 2x - 1 = 3x + 9 and x = -10.
6. 6. If y = \( \frac{-1}{2} \): 2(2x - 1) = -(x + 3), so 5x = -1 and x = \( \frac{-1}{5} \).
7. 7. Neither value is -3 or \( \frac{1}{2} \), so both are valid. Hence, x = -10 or x = \( \frac{-1}{5} \).

Teacher's Note:
a) Multiply by the LCD to remove the denominators, then simplify to a quadratic.
b) Always compare the roots with the values excluded in the question.

9. 5^(x+1) + 5^(2-x) = 5³ + 1

Answer:
1. 1. Write 5^x+1 = 5 × 5^x and 5^2-x = \( \frac{25}{5^x} \). Put t = 5^x; also 5³ + 1 = 126.
2. 2. The equation becomes 5t + \( \frac{25}{t} \) = 126, so 5t² - 126t + 25 = 0.
3. 3. Splitting the middle term: 5t² - 125t - t + 25 = 0, so 5t(t - 25) - 1(t - 25) = 0.
4. 4. (5t - 1)(t - 25) = 0, so t = \( \frac{1}{5} \) or t = 25.
5. 5. If 5^x = 25 = 5², then x = 2.
6. 6. If 5^x = \( \frac{1}{5} \) = 5^-1, then x = -1.
7. 7. Hence, x = 2 or x = -1.

Teacher's Note:
a) Write a^(x+1) as a × a^x so that every term uses the same power.
b) Reject any value of t that is not positive, since a power of a positive base is always positive.

10. 5x - \( \frac{35}{x} \) = 18, x ≠ 0

Answer:
1. Multiplying both sides by the LCD, x, to remove the denominators and simplifying: 5x² - 18x - 35 = 0.
2. Here a × c = 5 × (-35) = -175; two numbers with sum -18 and product -175 are 7 and -25.
3. Splitting the middle term: 5x² + 7x - 25x - 35 = 0.
4. Grouping: x(5x + 7) - 5(5x + 7) = 0.
5. (x - 5)(5x + 7) = 0.
6. So x - 5 = 0 or 5x + 7 = 0, which gives x = 5 or x = \( - \frac{7}{5} \).
7. Neither value makes a denominator zero, so both are valid. Hence, the roots are 5 and \( - \frac{7}{5} \).

11. 2^2x - 3.2^(x+2) + 32 = 0

Answer:
1. 1. Write 2^2x = (2^x)² and 2^x+2 = 4 × 2^x. Put t = 2^x.
2. 2. The equation becomes t² - 12t + 32 = 0.
3. 3. Splitting the middle term: t² - 4t - 8t + 32 = 0, so t(t - 4) - 8(t - 4) = 0.
4. 4. (t - 8)(t - 4) = 0, so t = 8 or t = 4.
5. 5. If 2^x = 8 = 2³, then x = 3.
6. 6. If 2^x = 4 = 2², then x = 2.
7. 7. Hence, x = 3 or x = 2.

Teacher's Note:
a) Put t = the repeated power to turn the equation into a quadratic in t.
b) Convert back using the base: if 2^x = 4, then x = 2.

12. 4^(x+1) + 4^(1-x) = 10

Answer:
1. 1. Write 4^x+1 = 4 × 4^x and 4^1-x = \( \frac{4}{4^x} \). Put t = 4^x.
2. 2. The equation becomes 4t + \( \frac{4}{t} \) = 10, so 4t² - 10t + 4 = 0, that is 2t² - 5t + 2 = 0.
3. 3. Splitting the middle term: 2t² - 4t - t + 2 = 0, so 2t(t - 2) - 1(t - 2) = 0.
4. 4. (2t - 1)(t - 2) = 0, so t = \( \frac{1}{2} \) or t = 2.
5. 5. If 4^x = 2, then 2^2x = 2¹ and x = \( \frac{1}{2} \).
6. 6. If 4^x = \( \frac{1}{2} \), then 2^2x = 2^-1 and x = \( \frac{-1}{2} \).
7. 7. Hence, x = \( \frac{1}{2} \) or x = \( \frac{-1}{2} \).

Teacher's Note:
a) Write a^(x+1) as a × a^x so that every term uses the same power.
b) Reject any value of t that is not positive, since a power of a positive base is always positive.

13. 3^(x+2) + 3^-x = 10

Answer:
1. 1. Write 3^x+2 = 9 × 3^x and 3^-x = \( \frac{1}{3^x} \). Put t = 3^x.
2. 2. The equation becomes 9t + \( \frac{1}{t} \) = 10, so 9t² - 10t + 1 = 0.
3. 3. Splitting the middle term: 9t² - 9t - t + 1 = 0, so 9t(t - 1) - 1(t - 1) = 0.
4. 4. (9t - 1)(t - 1) = 0, so t = \( \frac{1}{9} \) or t = 1.
5. 5. If 3^x = 1 = 3⁰, then x = 0.
6. 6. If 3^x = \( \frac{1}{9} \) = 3^-2, then x = -2.
7. 7. Hence, x = 0 or x = -2.

Teacher's Note:
a) Put t = the repeated power to turn the equation into a quadratic in t.
b) Convert back using the base: if 2^x = 4, then x = 2.

14. 10x - \( \frac{1}{x} \) = 3

Answer:
1. Multiplying both sides by the LCD, x, to remove the denominators and simplifying: 10x² - 3x - 1 = 0.
2. Here a × c = 10 × (-1) = -10; two numbers with sum -3 and product -10 are 2 and -5.
3. Splitting the middle term: 10x² + 2x - 5x - 1 = 0.
4. Grouping: 2x(5x + 1) - (5x + 1) = 0.
5. (2x - 1)(5x + 1) = 0.
6. So 2x - 1 = 0 or 5x + 1 = 0, which gives x = \( \frac{1}{2} \) or x = \( - \frac{1}{5} \).
7. Neither value makes a denominator zero, so both are valid. Hence, the roots are \( \frac{1}{2} \) and \( - \frac{1}{5} \).

Teacher's Note:
a) Reduce the equation to ax² + bx + c = 0 and divide by any common factor first.
b) Both roots are valid only if neither makes a denominator zero.

15. \( \frac{2}{x²} \) - \( \frac{5}{x} \) + 2 = 0

Answer:
1. Multiplying both sides by the LCD, x², to remove the denominators and simplifying: 2x² - 5x + 2 = 0.
2. Here a × c = 2 × 2 = 4; two numbers with sum -5 and product 4 are -1 and -4.
3. Splitting the middle term: 2x² - x - 4x + 2 = 0.
4. Grouping: x(2x - 1) - 2(2x - 1) = 0.
5. (x - 2)(2x - 1) = 0.
6. So x - 2 = 0 or 2x - 1 = 0, which gives x = 2 or x = \( \frac{1}{2} \).
7. Neither value makes a denominator zero, so both are valid. Hence, the roots are 2 and \( \frac{1}{2} \).

Teacher's Note:
a) Multiply by the LCD to remove the denominators, then simplify to a quadratic.
b) Always compare the roots with the values excluded in the question.

16. √3x² + 11x + 6√3 = 0

Answer:
1. 1. Here a × c = √3 × 6√3 = 18; two numbers with sum 11 and product 18 are 2 and 9.
2. 2. Splitting the middle term: √3x² + 2x + 9x + 6√3 = 0.
3. 3. Grouping: x(√3x + 2) + 3√3(√3x + 2) = 0.
4. 4. (x + 3√3)(√3x + 2) = 0.
5. 5. So x = -3√3 or x = \( -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3} \).
6. 6. Hence, the roots are -3√3 and \( -\frac{2\sqrt{3}}{3} \).

17. 4√3x² + 5x - 2√3 = 0

Answer:
1. 1. Here a × c = 4√3 × (-2√3) = -24; two numbers with sum 5 and product -24 are 8 and -3.
2. 2. Splitting the middle term: 4√3x² + 8x - 3x - 2√3 = 0.
3. 3. Grouping: 4x(√3x + 2) - √3(√3x + 2) = 0.
4. 4. (4x - √3)(√3x + 2) = 0.
5. 5. So x = \( \frac{\sqrt{3}}{4} \) or x = \( -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3} \).
6. 6. Hence, the roots are \( \frac{\sqrt{3}}{4} \) and \( -\frac{2\sqrt{3}}{3} \).

Teacher's Note:
a) Square roots such as √3 are treated like numbers when splitting the middle term.
b) Rationalise the denominator in the final roots.

18. 3√7x² + 4x - √7 = 0

Answer:
1. 1. Here a × c = 3√7 × (-√7) = -21; two numbers with sum 4 and product -21 are 7 and -3.
2. 2. Splitting the middle term: 3√7x² + 7x - 3x - √7 = 0.
3. 3. Grouping: √7x(3x + √7) - 1(3x + √7) = 0.
4. 4. (√7x - 1)(3x + √7) = 0.
5. 5. So x = \( \frac{1}{\sqrt{7}} = \frac{\sqrt{7}}{7} \) or x = \( -\frac{\sqrt{7}}{3} \).
6. 6. Hence, the roots are \( \frac{\sqrt{7}}{7} \) and \( -\frac{\sqrt{7}}{3} \).

19. √7x² - 6x - 13√7 = 0

Answer:
1. 1. Here a × c = √7 × (-13√7) = -91; two numbers with sum -6 and product -91 are 7 and -13.
2. 2. Splitting the middle term: √7x² + 7x - 13x - 13√7 = 0.
3. 3. Grouping: √7x(x + √7) - 13(x + √7) = 0.
4. 4. (√7x - 13)(x + √7) = 0.
5. 5. So x = \( \frac{13}{\sqrt{7}} = \frac{13\sqrt{7}}{7} \) or x = -√7.
6. 6. Hence, the roots are \( \frac{13\sqrt{7}}{7} \) and -√7.

Teacher's Note:
a) Square roots such as √3 are treated like numbers when splitting the middle term.
b) Rationalise the denominator in the final roots.

20. 4√6x² - 13x - 2√6 = 0

Answer:
1. 1. Here a × c = 4√6 × (-2√6) = -48; two numbers with sum -13 and product -48 are 3 and -16.
2. 2. Splitting the middle term: 4√6x² - 16x + 3x - 2√6 = 0.
3. 3. Grouping: 4x(√6x - 4) + \( \frac{\sqrt{6}}{2} \)(√6x - 4) = 0.
4. 4. \( \left(4x + \frac{\sqrt{6}}{2}\right) \)(√6x - 4) = 0, that is (8x + √6)(√6x - 4) = 0.
5. 5. So x = \( -\frac{\sqrt{6}}{8} \) or x = \( \frac{4}{\sqrt{6}} = \frac{2\sqrt{6}}{3} \).
6. 6. Hence, the roots are \( -\frac{\sqrt{6}}{8} \) and \( \frac{2\sqrt{6}}{3} \).

21. x² - (1 + √2)x + √2 = 0

Answer:
1. 1. Splitting the middle term: x² - x - √2x + √2 = 0.
2. 2. Grouping: x(x - 1) - √2(x - 1) = 0.
3. 3. (x - 1)(x - √2) = 0.
4. 4. Hence, x = 1 or x = √2.

Teacher's Note:
a) Square roots such as √3 are treated like numbers when splitting the middle term.
b) Rationalise the denominator in the final roots.

22. \( \left(\frac{4x - 3}{2x + 1}\right) \) - 10\( \left(\frac{2x + 1}{4x - 3}\right) \) = 3, \( \left(x \neq \frac{-1}{2}, \frac{3}{4}\right) \)

Answer:
1. 1. Put y = \( \frac{4x - 3}{2x + 1} \). Then the equation becomes y - \( \frac{10}{y} \) = 3.
2. 2. Multiplying by y: y² - 3y - 10 = 0.
3. 3. Splitting the middle term: y² - 5y + 2y - 10 = 0, so y(y - 5) + 2(y - 5) = 0.
4. 4. (y + 2)(y - 5) = 0, so y = 5 or y = -2.
5. 5. If y = 5: 4x - 3 = 5(2x + 1), so 4x - 3 = 10x + 5 and x = \( \frac{-4}{3} \).
6. 6. If y = -2: 4x - 3 = -2(2x + 1), so 8x = 1 and x = \( \frac{1}{8} \).
7. 7. Neither value is \( \frac{-1}{2} \) or \( \frac{3}{4} \), so both are valid. Hence, x = \( \frac{-4}{3} \) or x = \( \frac{1}{8} \).

23. \( \left(\frac{x}{x + 1}\right)^2 \) - 5\( \left(\frac{x}{x + 1}\right) \) + 6 = 0, (x ≠ -1)

Answer:
1. 1. Put y = \( \frac{x}{x + 1} \). The equation becomes y² - 5y + 6 = 0.
2. 2. Splitting the middle term: y² - 2y - 3y + 6 = 0, so y(y - 2) - 3(y - 2) = 0.
3. 3. (y - 3)(y - 2) = 0, so y = 3 or y = 2.
4. 4. If y = 2: x = 2(x + 1), so x = 2x + 2 and x = -2.
5. 5. If y = 3: x = 3(x + 1), so x = 3x + 3 and x = \( \frac{-3}{2} \).
6. 6. Neither value is -1, so both are valid. Hence, x = -2 or x = \( \frac{-3}{2} \).

Teacher's Note:
a) Reduce the equation to ax² + bx + c = 0 and divide by any common factor first.
b) Both roots are valid only if neither makes a denominator zero.

24. 2\( \left(\frac{2x - 1}{x + 3}\right) \) - 3\( \left(\frac{x + 3}{2x - 1}\right) \) = 5, x ≠ -3, \( \frac{1}{2} \)

Teacher's Note:
a) Multiply by the LCD to remove the denominators, then simplify to a quadratic.
b) Always compare the roots with the values excluded in the question.

25. 2\( \left(\frac{x - 1}{x + 3}\right) \) - 7\( \left(\frac{x + 3}{x - 1}\right) \) = 5, (x ≠ -3, 1)

Answer:
1. 1. Put y = \( \frac{x - 1}{x + 3} \). Then \( \frac{x + 3}{x - 1} \) = \( \frac{1}{y} \) and the equation becomes 2y - \( \frac{7}{y} \) = 5.
2. 2. Multiplying by y: 2y² - 5y - 7 = 0.
3. 3. Splitting the middle term: 2y² - 7y + 2y - 7 = 0, so y(2y - 7) + 1(2y - 7) = 0.
4. 4. (y + 1)(2y - 7) = 0, so y = -1 or y = \( \frac{7}{2} \).
5. 5. If y = -1: x - 1 = -(x + 3), so 2x = -2 and x = -1.
6. 6. If y = \( \frac{7}{2} \): 2(x - 1) = 7(x + 3), so 2x - 2 = 7x + 21 and x = \( \frac{-23}{5} \).
7. 7. Neither value is -3 or 1, so both are valid. Hence, x = -1 or x = \( \frac{-23}{5} \).

26. \( \frac{a}{x - b} \) + \( \frac{b}{x - a} \) = 2, (x ≠ a, b)

Answer:
1. 1. Taking the LCM: a(x - a) + b(x - b) = 2(x - a)(x - b).
2. 2. ax - a² + bx - b² = 2x² - 2(a + b)x + 2ab.
3. 3. Collecting terms: 2x² - 3(a + b)x + (a + b)² = 0.
4. 4. Splitting the middle term: 2x² - 2(a + b)x - (a + b)x + (a + b)² = 0.
5. 5. Grouping: 2x(x - (a + b)) - (a + b)(x - (a + b)) = 0.
6. 6. (2x - (a + b))(x - (a + b)) = 0.
7. 7. Hence, x = \( \frac{a + b}{2} \) or x = a + b.

Teacher's Note:
a) Treat a and b as constants and factorise exactly as with numbers.
b) Check by substituting a root back in the equation.

27. \( \frac{a}{ax - 1} \) + \( \frac{b}{bx - 1} \) = a + b, \( \left(x \neq \frac{1}{a}, \frac{1}{b}\right) \)

Answer:
1. 1. Taking the LCM: a(bx - 1) + b(ax - 1) = (a + b)(ax - 1)(bx - 1).
2. 2. Left side: 2abx - (a + b). Right side: (a + b)(abx² - (a + b)x + 1).
3. 3. Collecting terms: ab(a + b)x² - (a² + 4ab + b²)x + 2(a + b) = 0.
4. 4. Splitting the middle term: ab(a + b)x² - 2abx - (a + b)²x + 2(a + b) = 0.
5. 5. Grouping: abx((a + b)x - 2) - (a + b)((a + b)x - 2) = 0.
6. 6. ((a + b)x - 2)(abx - (a + b)) = 0.
7. 7. Hence, x = \( \frac{2}{a + b} \) or x = \( \frac{a + b}{ab} \).

28. \( \frac{x + 3}{x - 2} \) - \( \frac{1 - x}{x} \) = \( \frac{17}{4} \), (x ≠ 0, 2)

Answer:
1. Multiplying both sides by the LCD, 4x(x - 2), to remove the denominators and simplifying: 9x² - 34x - 8 = 0.
2. Here a × c = 9 × (-8) = -72; two numbers with sum -34 and product -72 are 2 and -36.
3. Splitting the middle term: 9x² + 2x - 36x - 8 = 0.
4. Grouping: x(9x + 2) - 4(9x + 2) = 0.
5. (x - 4)(9x + 2) = 0.
6. So x - 4 = 0 or 9x + 2 = 0, which gives x = 4 or x = \( - \frac{2}{9} \).
7. Neither value makes a denominator zero, so both are valid. Hence, the roots are 4 and \( - \frac{2}{9} \).

Teacher's Note:
a) Reduce the equation to ax² + bx + c = 0 and divide by any common factor first.
b) Both roots are valid only if neither makes a denominator zero.

29. \( \frac{2x}{x - 4} \) + \( \frac{2x - 5}{x - 3} \) = \( \frac{25}{3} \), (x ≠ 4, 3)

Answer:
1. Multiplying both sides by the LCD, 3(x - 4)(x - 3), to remove the denominators and simplifying: 13x² - 118x + 240 = 0.
2. Here a × c = 13 × 240 = 3120; two numbers with sum -118 and product 3120 are -40 and -78.
3. Splitting the middle term: 13x² - 40x - 78x + 240 = 0.
4. Grouping: x(13x - 40) - 6(13x - 40) = 0.
5. (x - 6)(13x - 40) = 0.
6. So x - 6 = 0 or 13x - 40 = 0, which gives x = 6 or x = \( \frac{40}{13} \).
7. Neither value makes a denominator zero, so both are valid. Hence, the roots are 6 and \( \frac{40}{13} \).

Teacher's Note:
a) Multiply by the LCD to remove the denominators, then simplify to a quadratic.
b) Always compare the roots with the values excluded in the question.

30. \( \frac{1}{x - 3} \) - \( \frac{1}{x + 5} \) = \( \frac{1}{6} \), (x ≠ 3, -5)

Answer:
1. Multiplying both sides by the LCD, 6(x - 3)(x + 5), to remove the denominators and simplifying: x² + 2x - 63 = 0.
2. Here a × c = 1 × (-63) = -63; two numbers with sum 2 and product -63 are -7 and 9.
3. Splitting the middle term: x² - 7x + 9x - 63 = 0.
4. Grouping: x(x - 7) + 9(x - 7) = 0.
5. (x + 9)(x - 7) = 0.
6. So x + 9 = 0 or x - 7 = 0, which gives x = -9 or x = 7.
7. Neither value makes a denominator zero, so both are valid. Hence, the roots are -9 and 7.

31. \( \frac{1}{x - 2} \) + \( \frac{2}{x - 1} \) = \( \frac{6}{x} \), (x ≠ 2, 1)

Answer:
1. Multiplying both sides by the LCD, x(x - 2)(x - 1), to remove the denominators and simplifying: 3x² - 13x + 12 = 0.
2. Here a × c = 3 × 12 = 36; two numbers with sum -13 and product 36 are -4 and -9.
3. Splitting the middle term: 3x² - 4x - 9x + 12 = 0.
4. Grouping: x(3x - 4) - 3(3x - 4) = 0.
5. (x - 3)(3x - 4) = 0.
6. So x - 3 = 0 or 3x - 4 = 0, which gives x = 3 or x = \( \frac{4}{3} \).
7. Neither value makes a denominator zero, so both are valid. Hence, the roots are 3 and \( \frac{4}{3} \).

Teacher's Note:
a) Reduce the equation to ax² + bx + c = 0 and divide by any common factor first.
b) Both roots are valid only if neither makes a denominator zero.

32. \( \frac{1}{x + 4} \) - \( \frac{1}{x - 7} \) = \( \frac{11}{30} \), (x ≠ -4, 7)

Answer:
1. Multiplying both sides by the LCD, 30(x - 7)(x + 4), to remove the denominators and simplifying: x² - 3x + 2 = 0.
2. Here a × c = 1 × 2 = 2; two numbers with sum -3 and product 2 are -1 and -2.
3. Splitting the middle term: x² - x - 2x + 2 = 0.
4. Grouping: x(x - 1) - 2(x - 1) = 0.
5. (x - 2)(x - 1) = 0.
6. So x - 2 = 0 or x - 1 = 0, which gives x = 2 or x = 1.
7. Neither value makes a denominator zero, so both are valid. Hence, the roots are 2 and 1.

Teacher's Note:
a) Multiply by the LCD to remove the denominators, then simplify to a quadratic.
b) Always compare the roots with the values excluded in the question.

33. \( \frac{1}{x - 2} \) + \( \frac{1}{x - 4} \) = \( \frac{4}{3} \), (x ≠ 2, 4)

Answer:
1. Multiplying both sides by the LCD, 3(x - 4)(x - 2), to remove the denominators and simplifying: 2x² - 15x + 25 = 0.
2. Here a × c = 2 × 25 = 50; two numbers with sum -15 and product 50 are -5 and -10.
3. Splitting the middle term: 2x² - 5x - 10x + 25 = 0.
4. Grouping: x(2x - 5) - 5(2x - 5) = 0.
5. (x - 5)(2x - 5) = 0.
6. So x - 5 = 0 or 2x - 5 = 0, which gives x = 5 or x = \( \frac{5}{2} \).
7. Neither value makes a denominator zero, so both are valid. Hence, the roots are 5 and \( \frac{5}{2} \).

34. \( \frac{x - 3}{x + 3} \) - \( \frac{x + 3}{x - 3} \) = 6\( \frac{6}{7} \), (x ≠ -3, 3)

Answer:
1. Writing 6\( \frac{6}{7} \) = \( \frac{48}{7} \) and cross-multiplying: 7[(x - 3)² - (x + 3)²] = 48(x + 3)(x - 3), so -84x = 48x² - 432, which gives 4x² + 7x - 36 = 0.
2. Here a × c = 4 × (-36) = -144; two numbers with sum 7 and product -144 are -9 and 16.
3. Splitting the middle term: 4x² - 9x + 16x - 36 = 0.
4. Grouping: x(4x - 9) + 4(4x - 9) = 0.
5. (x + 4)(4x - 9) = 0.
6. So x + 4 = 0 or 4x - 9 = 0, which gives x = -4 or x = \( \frac{9}{4} \).
7. Neither value makes a denominator zero, so both are valid. Hence, the roots are -4 and \( \frac{9}{4} \).

Teacher's Note:
a) Reduce the equation to ax² + bx + c = 0 and divide by any common factor first.
b) Both roots are valid only if neither makes a denominator zero.

35. \( \frac{2x}{x - 3} \) + \( \frac{1}{2x + 3} \) + \( \frac{3x + 9}{(x - 3)(2x + 3)} \) = 0

Answer:
1. Multiplying by (x - 3)(2x + 3): 2x(2x + 3) + (x - 3) + (3x + 9) = 0, so 4x² + 10x + 6 = 0, that is 2x² + 5x + 3 = 0.
2. Here a × c = 2 × 3 = 6; two numbers with sum 5 and product 6 are 2 and 3.
3. Splitting the middle term: 2x² + 2x + 3x + 3 = 0.
4. Grouping: 2x(x + 1) + 3(x + 1) = 0.
5. (2x + 3)(x + 1) = 0.
6. So 2x + 3 = 0 or x + 1 = 0, which gives x = \( - \frac{3}{2} \) or x = -1.
7. The value x = \( - \frac{3}{2} \) makes a denominator zero, so it is rejected.
8. Hence, the solution is x = -1.

Teacher's Note:
a) Multiply by the LCD to remove the denominators, then simplify to a quadratic.
b) Always compare the roots with the values excluded in the question.
c) Check every root against the denominators: x = -3/2 makes 2x + 3 zero, so it must be rejected.

36. x = \( \frac{1}{2 - \frac{1}{2 - \frac{1}{2 - x}}} \), x ≠ 2

Answer:
1. 1. Innermost part: 2 - \( \frac{1}{2 - x} \) = \( \frac{4 - 2x - 1}{2 - x} \) = \( \frac{3 - 2x}{2 - x} \).
2. 2. Next: 2 - \( \frac{2 - x}{3 - 2x} \) = \( \frac{6 - 4x - 2 + x}{3 - 2x} \) = \( \frac{4 - 3x}{3 - 2x} \).
3. 3. So the equation is x = \( \frac{3 - 2x}{4 - 3x} \).
4. 4. Cross-multiplying: x(4 - 3x) = 3 - 2x, so 4x - 3x² = 3 - 2x.
5. 5. 3x² - 6x + 3 = 0, that is x² - 2x + 1 = 0.
6. 6. (x - 1)² = 0, so x = 1.
7. 7. For x = 1 no denominator is zero, so x = 1 is the solution (equal roots).

37. 4x² - 2(a² + b²)x + a²b² = 0

Answer:
1. 1. Here a × c = 4a²b²; we need two terms with sum -2(a² + b²) and product 4a²b²; they are -2a² and -2b².
2. 2. Splitting the middle term: 4x² - 2a²x - 2b²x + a²b² = 0.
3. 3. Grouping: 2x(2x - a²) - b²(2x - a²) = 0.
4. 4. (2x - a²)(2x - b²) = 0.
5. 5. Hence, x = \( \frac{a²}{2} \) or x = \( \frac{b²}{2} \).

Teacher's Note:
a) Treat a and b as constants and factorise exactly as with numbers.
b) Check by substituting a root back in the equation.

38. 9x² - 9(a + b)x + (2a² + 5ab + 2b²) = 0

Answer:
1. 1. Factorise the constant: 2a² + 5ab + 2b² = (2a + b)(a + 2b).
2. 2. We need two terms with sum -9(a + b) and product 9(2a + b)(a + 2b); they are -3(2a + b) and -3(a + 2b).
3. 3. Splitting the middle term: 9x² - 3(2a + b)x - 3(a + 2b)x + (2a + b)(a + 2b) = 0.
4. 4. Grouping: 3x(3x - (2a + b)) - (a + 2b)(3x - (2a + b)) = 0.
5. 5. (3x - (2a + b))(3x - (a + 2b)) = 0.
6. 6. Hence, x = \( \frac{2a + b}{3} \) or x = \( \frac{a + 2b}{3} \).

39. 4x² - 4a²x + (a⁴ - b⁴) = 0

Answer:
1. 1. Factorise the constant: a⁴ - b⁴ = (a² + b²)(a² - b²).
2. 2. We need two terms with sum -4a² and product 4(a² + b²)(a² - b²); they are -2(a² + b²) and -2(a² - b²).
3. 3. Splitting the middle term: 4x² - 2(a² + b²)x - 2(a² - b²)x + (a² + b²)(a² - b²) = 0.
4. 4. Grouping: 2x(2x - (a² + b²)) - (a² - b²)(2x - (a² + b²)) = 0.
5. 5. (2x - (a² + b²))(2x - (a² - b²)) = 0.
6. 6. Hence, x = \( \frac{a² + b²}{2} \) or x = \( \frac{a² - b²}{2} \).

Teacher's Note:
a) Treat a and b as constants and factorise exactly as with numbers.
b) Check by substituting a root back in the equation.

40. x² + (\( \frac{a + b}{a} \) + \( \frac{a}{a + b} \))x + 1 = 0

Answer:
1. 1. Splitting the middle term as the two given parts: x² + \( \frac{a + b}{a} \)x + \( \frac{a}{a + b} \)x + 1 = 0.
2. 2. Grouping: x(x + \( \frac{a + b}{a} \)) + \( \frac{a}{a + b} \)(x + \( \frac{a + b}{a} \)) = 0, since \( \frac{a}{a + b} \) × \( \frac{a + b}{a} \) = 1.
3. 3. (x + \( \frac{a + b}{a} \))(x + \( \frac{a}{a + b} \)) = 0.
4. 4. Hence, x = \( -\frac{a + b}{a} \) or x = \( -\frac{a}{a + b} \).

41. x² + x - (a + 1)(a + 2) = 0

Answer:
1. 1. We need two terms with sum 1 and product -(a + 1)(a + 2); they are (a + 2) and -(a + 1).
2. 2. Splitting the middle term: x² + (a + 2)x - (a + 1)x - (a + 1)(a + 2) = 0.
3. 3. Grouping: x(x + a + 2) - (a + 1)(x + a + 2) = 0.
4. 4. (x + a + 2)(x - a - 1) = 0.
5. 5. Hence, x = -(a + 2) or x = a + 1.

Teacher's Note:
a) Treat a and b as constants and factorise exactly as with numbers.
b) Check by substituting a root back in the equation.

42. x² + 3x - (a² + a - 2) = 0

Answer:
1. 1. Factorise the constant: a² + a - 2 = (a + 2)(a - 1).
2. 2. We need two terms with sum 3 and product -(a + 2)(a - 1); they are (a + 2) and -(a - 1), since (a + 2) - (a - 1) = 3.
3. 3. Splitting the middle term: x² + (a + 2)x - (a - 1)x - (a + 2)(a - 1) = 0.
4. 4. Grouping: x(x + a + 2) - (a - 1)(x + a + 2) = 0.
5. 5. (x + a + 2)(x - a + 1) = 0.
6. 6. Hence, x = -(a + 2) or x = a - 1.

43. a²b²x² + b²x - a²x - 1 = 0

Answer:
1. 1. Grouping the terms: b²x(a²x + 1) - 1(a²x + 1) = 0.
2. 2. (a²x + 1)(b²x - 1) = 0.
3. 3. Hence, x = \( -\frac{1}{a^2} \) or x = \( \frac{1}{b^2} \).

Teacher's Note:
a) Treat a and b as constants and factorise exactly as with numbers.
b) Check by substituting a root back in the equation.

44. x + \( \frac{1}{x} \) = 25\( \frac{1}{25} \)

Answer:
1. 1. Writing 25\( \frac{1}{25} \) = \( \frac{626}{25} \) and multiplying by 25x: 25x² + 25 = 626x, so 25x² - 626x + 25 = 0.
2. 2. Here a × c = 625; two numbers with sum -626 and product 625 are -1 and -625.
3. 3. Splitting the middle term: 25x² - x - 625x + 25 = 0.
4. 4. Grouping: x(25x - 1) - 25(25x - 1) = 0.
5. 5. (x - 25)(25x - 1) = 0.
6. 6. Hence, x = 25 or x = \( \frac{1}{25} \).

Teacher's Note:
a) After factorising, each bracket gives one root; do not forget to change the sign of the constant.
b) A quick check: sum of roots = -b/a and product = c/a.

45. (x - 3)(x - 4) = \( \frac{34}{(33)²} \)

Answer:
1. 1. Expanding: x² - 7x + 12 = \( \frac{34}{1089} \). Multiplying by 1089: 1089x² - 7623x + 13034 = 0.
2. 2. Put t = 33x. Then t² - 231t + 13034 = 0.
3. 3. Two numbers with sum 231 and product 13034 are 98 and 133.
4. 4. Splitting the middle term: t² - 98t - 133t + 13034 = 0, so t(t - 98) - 133(t - 98) = 0.
5. 5. (t - 133)(t - 98) = 0, so t = 133 or t = 98.
6. 6. Since t = 33x, x = \( \frac{133}{33} \) or x = \( \frac{98}{33} \).

Teacher's Note:
a) Split the middle term so that the two parts add up to b and multiply to a × c.
b) Equate each factor to zero to get both roots.

46. x² + (a + \( \frac{1}{a} \))x + 1 = 0

Answer:
1. 1. Splitting the middle term as the two given parts: x² + ax + \( \frac{1}{a} \)x + 1 = 0.
2. 2. Grouping: x(x + a) + \( \frac{1}{a} \)(x + a) = 0, since \( \frac{1}{a} \) × a = 1.
3. 3. (x + a)(x + \( \frac{1}{a} \)) = 0.
4. 4. Hence, x = -a or x = \( -\frac{1}{a} \).

47. (a + b)²x² - 4abx - (a - b)² = 0

Answer:
1. 1. Since (a - b)² - (a + b)² = -4ab, split the middle term: -4abx = (a - b)²x - (a + b)²x.
2. 2. (a + b)²x² + (a - b)²x - (a + b)²x - (a - b)² = 0.
3. 3. Grouping: x((a + b)²x + (a - b)²) - 1((a + b)²x + (a - b)²) = 0.
4. 4. (x - 1)((a + b)²x + (a - b)²) = 0.
5. 5. Hence, x = 1 or x = \( -\frac{(a - b)^2}{(a + b)^2} \).

Teacher's Note:
a) Treat a and b as constants and factorise exactly as with numbers.
b) Check by substituting a root back in the equation.

48. 7x + \( \frac{3}{x} \) = 35\( \frac{3}{5} \)

Answer:
1. Writing 35\( \frac{3}{5} \) = \( \frac{178}{5} \) and multiplying by 5x: 35x² + 15 = 178x, so 35x² - 178x + 15 = 0.
2. Here a × c = 35 × 15 = 525; two numbers with sum -178 and product 525 are -3 and -175.
3. Splitting the middle term: 35x² - 3x - 175x + 15 = 0.
4. Grouping: x(35x - 3) - 5(35x - 3) = 0.
5. (x - 5)(35x - 3) = 0.
6. So x - 5 = 0 or 35x - 3 = 0, which gives x = 5 or x = \( \frac{3}{35} \).
7. Neither value makes a denominator zero, so both are valid. Hence, the roots are 5 and \( \frac{3}{35} \).

Teacher's Note:
a) Reduce the equation to ax² + bx + c = 0 and divide by any common factor first.
b) Both roots are valid only if neither makes a denominator zero.

49. \( \frac{x - a}{x - b} \) + \( \frac{x - b}{x - a} \) = \( \frac{a}{b} \) + \( \frac{b}{a} \)

Answer:
1. 1. Taking the LCM on both sides: \( \frac{(x - a)^2 + (x - b)^2}{(x - a)(x - b)} = \frac{a^2 + b^2}{ab} \).
2. 2. Cross-multiplying: ab[(x - a)² + (x - b)²] = (a² + b²)(x - a)(x - b).
3. 3. ab[2x² - 2(a + b)x + a² + b²] = (a² + b²)[x² - (a + b)x + ab].
4. 4. The constant terms ab(a² + b²) cancel on both sides, leaving -(a - b)²x² + (a + b)(a - b)²x = 0.
5. 5. (a - b)² x (a + b - x) = 0.
6. 6. Since a ≠ b, x = 0 or x = a + b.

50. (x - 5)(x - 6) = \( \frac{25}{(24)²} \)

Answer:
1. 1. Expanding: x² - 11x + 30 = \( \frac{25}{576} \). Multiplying by 576: 576x² - 6336x + 17255 = 0.
2. 2. Put t = 24x. Then t² - 264t + 17255 = 0.
3. 3. Two numbers with sum 264 and product 17255 are 119 and 145.
4. 4. Splitting the middle term: t² - 119t - 145t + 17255 = 0, so t(t - 119) - 145(t - 119) = 0.
5. 5. (t - 145)(t - 119) = 0, so t = 145 or t = 119.
6. 6. Since t = 24x, x = \( \frac{145}{24} \) or x = \( \frac{119}{24} \).

5. y²– 4y + 3 = 0

6. x² – 21x + 108 = 0

12. 7x²+ 49x + 84 = 0

13. m²+ 17mn – 84n²= 0

14. 5x²+ 16x + 3 = 0

15. 6x²+ 17x +12 = 0

16. 9x² + 18x + 8 = 0

17. 14x²+ 9x + 1 = 0

18. 2x² + 3x – 90 = 0

19. 2x²+ 11x – 21 = 0

20. 3x² – 14x + 8 = 0

21. 18x²+ 3x – 10 = 0

22. 15x²+ 2x – 8 = 0

23. 6x²+ 11x – 10 = 0

24. 30x²+ 7x – 15 = 0

25. 24x²– 41x + 12 = 0

27. 6x²+ 11x – 10 = 0

28. 10x²– 9x – 7 =0

29. 5x²– 16x – 21 = 0

31. 15x²– x – 28 = 0

32. 8a²– 27ab + 9b²= 0

33. 5x²+ 33xy – 14y²= 0

34. 3x³ – x²– 10x = 0

36. x²+ 5x – 24 = 0

37. x² – 4x – 21 = 0

38. 6x²+ 7x – 3 = 0

40. 9x²– 22x + 8 = 0

41. 6x²+40=31x

42. 36x²-12ax+(a²-b²)=0

43. 8x²-22x-21=0

Facts that Matter

Quadratic Equations: An equation of the form \( ax^2 + bx + c = 0 \), where \( a, b, c \) are real numbers, and \( a \neq 0 \), is called a quadratic equation in the variable \( x \).
Examples: \( x^2 - 3x + 4 = 0 \), \( \sqrt{2}y^2 + \frac{3}{\sqrt{2}}y - 2 = 0 \), \( 2x^2 - 149 = 0 \).
Roots of a Quadratic Equation:
- A value \( x = \alpha \) is said to be a root of \( ax^2 + bx + c = 0 \) if it satisfies the equation.
- Example: 2 is a root of \( 3x^2 - x - 10 = 0 \) because \( 3(2)^2 - 2 - 10 = 0 \).
- A quadratic equation has at most two roots (zeroes). Either it has no zero, or two zeroes.
- We can solve a quadratic equation by factorisation.

EXERCISE 4.1

Question. Check whether the following are quadratic equations: \( (x + 1)^2 = 2(x - 3) \)
Answer: We have:
\( (x + 1)^2 = 2(x - 3) \)
\( \Rightarrow x^2 + 2x + 1 = 2x - 6 \)
\( \Rightarrow x^2 + 2x + 1 - 2x + 6 = 0 \)
\( \Rightarrow x^2 + 7 = 0 \)
Since \( x^2 + 7 \) is a quadratic polynomial
\( \therefore (x + 1)^2 = 2(x - 3) \) is a quadratic equation.

Question. Check whether the following are quadratic equations: \( x^2 - 2x = (-2)(3 - x) \)
Answer: We have:
\( x^2 - 2x = (-2)(3 - x) \)
\( \Rightarrow x^2 - 2x = -6 + 2x \)
\( \Rightarrow x^2 - 2x - 2x + 6 = 0 \)
\( \Rightarrow x^2 - 4x + 6 = 0 \)
Since \( x^2 - 4x + 6 \) is a quadratic polynomial
\( \therefore x^2 - 2x = (-2)(3 - x) \) is a quadratic equation.

Question. Check whether the following are quadratic equations: \( (x - 2)(x + 1) = (x - 1)(x + 3) \)
Answer: We have:
\( (x - 2)(x + 1) = (x - 1)(x + 3) \)
\( \Rightarrow x^2 - x - 2 = x^2 + 2x - 3 \)
\( \Rightarrow x^2 - x - 2 - x^2 - 2x + 3 = 0 \)
\( \Rightarrow -3x + 1 = 0 \)
Since \( -3x + 1 \) is a linear polynomial
\( \therefore (x - 2)(x + 1) = (x - 1)(x + 3) \) is not quadratic equation.

Question. Check whether the following are quadratic equations: \( (x - 3)(2x + 1) = x(x + 5) \)
Answer: We have:
\( (x - 3)(2x + 1) = x(x + 5) \)
\( \Rightarrow 2x^2 + x - 6x - 3 = x^2 + 5x \)
\( \Rightarrow 2x^2 - 5x - 3 - x^2 - 5x = 0 \)
\( \Rightarrow x^2 - 10x - 3 = 0 \)
Since \( x^2 - 10x - 3 \) is a quadratic polynomial
\( \therefore (x - 3)(2x + 1) = x(x + 5) \) is a quadratic equation.

Question. Check whether the following are quadratic equations: \( (2x - 1)(x - 3) = (x + 5)(x - 1) \)
Answer: We have:
\( (2x - 1)(x - 3) = (x + 5)(x - 1) \)
\( \Rightarrow 2x^2 - 6x - x + 3 = x^2 - x + 5x - 5 \)
\( \Rightarrow 2x^2 - x^2 - 6x - x + x - 5x + 3 + 5 = 0 \)
\( \Rightarrow x^2 - 11x + 8 = 0 \)
Since \( x^2 - 11x + 8 \) is a quadratic polynomial
\( \therefore (2x - 1)(x - 3) = (x + 5)(x - 1) \) is a quadratic equation.

Question. Check whether the following are quadratic equations: \( x^2 + 3x + 1 = (x - 2)^2 \)
Answer: We have:
\( x^2 + 3x + 1 = (x - 2)^2 \)
\( \Rightarrow x^2 + 3x + 1 = x^2 - 4x + 4 \)
\( \Rightarrow x^2 + 3x + 1 - x^2 + 4x - 4 = 0 \)
\( \Rightarrow 7x - 3 = 0 \)
Since \( 7x - 3 \) is a linear polynomial.
\( \therefore x^2 + 3x + 1 = (x - 2)^2 \) is not a quadratic equation.

Question. Check whether the following are quadratic equations: \( (x + 2)^3 = 2x(x^2 - 1) \)
Answer: We have:
\( (x + 2)^3 = 2x(x^2 - 1) \)
\( \Rightarrow x^3 + 3x^2(2) + 3x(2)^2 + (2)^3 = 2x^3 - 2x \)
\( \Rightarrow x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x \)
\( \Rightarrow x^3 + 6x^2 + 12x + 8 - 2x^3 + 2x = 0 \)
\( \Rightarrow -x^3 + 6x^2 + 14x + 8 = 0 \)
Since \( -x^3 + 6x^2 + 14x + 8 \) is a polynomial of degree 3
\( \therefore (x + 2)^3 = 2x(x^2 - 1) \) is not a quadratic equation.

Question. Check whether the following are quadratic equations: \( x^3 - 4x^2 - x + 1 = (x - 2)^3 \)
Answer: We have:
\( x^3 - 4x^2 - x + 1 = (x - 2)^3 \)
\( \Rightarrow x^3 - 4x^2 - x + 1 = x^3 + 3x^2(-2) + 3x(-2)^2 + (-2)^3 \)
\( \Rightarrow x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8 \)
\( \Rightarrow x^3 - 4x^2 - x + 1 - x^3 + 6x^2 - 12x + 8 = 0 \)
\( \Rightarrow 2x^2 - 13x + 9 = 0 \)
Since \( 2x^2 - 13x + 9 \) is a quadratic polynomial
\( \therefore x^3 - 4x^2 - x + 1 = (x - 2)^3 \) is a quadratic equation.

Question. Represent the following situation in the form of quadratic equations: The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Answer: Let the breadth = \( x \) metres
\( \therefore \) Length = \( 2(\text{Breadth}) + 1 = 2x + 1 \) metres
Since Length \( \times \) Breadth = Area
\( \therefore (2x + 1) \times x = 528 \)
\( \Rightarrow 2x^2 + x = 528 \)
\( \Rightarrow 2x^2 + x - 528 = 0 \)
Thus, the required quadratic equation is \( 2x^2 + x - 528 = 0 \).

Question. Represent the following situation in the form of quadratic equations: The product of two consecutive positive integers is 306. We need to find the integers.
Answer: Let the two consecutive numbers be \( x \) and \( (x + 1) \).
Given product of the numbers = 306
\( \therefore x(x + 1) = 306 \)
\( \Rightarrow x^2 + x = 306 \)
\( \Rightarrow x^2 + x - 306 = 0 \)
Thus, the required quadratic equation is \( x^2 + x - 306 = 0 \).

Question. Represent the following situation in the form of quadratic equations: Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Answer: Let the present age = \( x \)
\( \therefore \) Mother’s age = \( (x + 26) \) years
After 3 years:
Rohan's age = \( (x + 3) \) years
Mother’s age = \( [(x + 26) + 3] = (x + 29) \) years
According to the condition,
\( (x + 3) \times (x + 29) = 360 \)
\( \Rightarrow x^2 + 29x + 3x + 87 = 360 \)
\( \Rightarrow x^2 + 32x + 87 - 360 = 0 \)
\( \Rightarrow x^2 + 32x - 273 = 0 \)
Thus, the required quadratic equation is \( x^2 + 32x - 273 = 0 \).

Question. Represent the following situation in the form of quadratic equations: A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Answer: Let the speed of the train = \( u \) km/hr
Distance covered = 480 km
Time taken = Distance \( \div \) Speed = \( \frac{480}{u} \) hours
In second case, Speed = \( (u - 8) \) km/hour
\( \therefore \) Time taken = \( \frac{480}{u - 8} \) hours
According to the condition, \( \frac{480}{u - 8} - \frac{480}{u} = 3 \)
\( \Rightarrow 480u - 480(u - 8) = 3u(u - 8) \)
\( \Rightarrow 480u - 480u + 3840 = 3u^2 - 24u \)
\( \Rightarrow 3u^2 - 24u - 3840 = 0 \)
\( \Rightarrow u^2 - 8u - 1280 = 0 \)
Thus, the required quadratic equation is \( u^2 - 8u - 1280 = 0 \).

Question. Find the roots of the following quadratic equation by factorisation: \( x^2 - 3x - 10 = 0 \)
Answer: We have:
\( x^2 - 3x - 10 = 0 \)
\( \Rightarrow x^2 - 5x + 2x - 10 = 0 \)
\( \Rightarrow x(x - 5) + 2(x - 5) = 0 \)
\( \Rightarrow (x - 5)(x + 2) = 0 \)
Either \( x - 5 = 0 \Rightarrow x = 5 \)
or \( x + 2 = 0 \Rightarrow x = -2 \)
Thus, the required roots are \( x = 5 \) and \( x = -2 \).

Question. Find the roots of the following quadratic equation by factorisation: \( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \)
Answer: We have:
\( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \)
\( \Rightarrow \sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = 0 \)
\( \Rightarrow \sqrt{2}x^2 + (\sqrt{2} \cdot \sqrt{2})x + 5x + 5\sqrt{2} = 0 \)
\( \Rightarrow \sqrt{2}x[x + \sqrt{2}] + 5[x + \sqrt{2}] = 0 \)
\( \Rightarrow (x + \sqrt{2})(\sqrt{2}x + 5) = 0 \)
Either \( x + \sqrt{2} = 0 \Rightarrow x = -\sqrt{2} \)
or \( \sqrt{2}x + 5 = 0 \Rightarrow x = \frac{-5}{\sqrt{2}} \)
Thus, the required roots are \( x = -\sqrt{2} \) and \( x = \frac{-5}{\sqrt{2}} \).

Question. Find the roots of the following quadratic equation by factorisation: \( 2x^2 - x + \frac{1}{8} = 0 \)
Answer: We have:
\( 2x^2 - x + \frac{1}{8} = 0 \)
\( \Rightarrow 16x^2 - 8x + 1 = 0 \)
\( \Rightarrow 16x^2 - 4x - 4x + 1 = 0 \)
\( \Rightarrow 4x(4x - 1) - 1(4x - 1) = 0 \)
\( \Rightarrow (4x - 1)(4x - 1) = 0 \)
\( \Rightarrow x = \frac{1}{4} \) and \( x = \frac{1}{4} \)
Thus, the required roots are \( x = \frac{1}{4} \) and \( x = \frac{1}{4} \).

Question. Find the roots of the following quadratic equation by factorisation: \( 100x^2 - 20x + 1 = 0 \)
Answer: We have:
\( 100x^2 - 20x + 1 = 0 \)
\( \Rightarrow 100x^2 - 10x - 10x + 1 = 0 \)
\( \Rightarrow 10x(10x - 1) - 1(10x - 1) = 0 \)
\( \Rightarrow (10x - 1)(10x - 1) = 0 \)
\( \Rightarrow x = \frac{1}{10} \) and \( x = \frac{1}{10} \)
Thus, the required roots are \( x = \frac{1}{10} \) and \( x = \frac{1}{10} \).

Question. Solve the problem: \( x^2 - 45x + 324 = 0 \)
Answer: We have:
\( x^2 - 45x + 324 = 0 \)
\( \Rightarrow x^2 - 9x - 36x + 324 = 0 \) [Since \( (-9) \times (-36) = 324 \) and \( (-9) + (-36) = -45 \)]
\( \Rightarrow x(x - 9) - 36(x - 9) = 0 \)
\( \Rightarrow (x - 9)(x - 36) = 0 \)
Either \( x - 9 = 0 \Rightarrow x = 9 \)
or \( x - 36 = 0 \Rightarrow x = 36 \)
Thus, \( x = 9 \) and \( x = 36 \).

Question. Solve the problem: \( x^2 - 55x + 750 = 0 \)
Answer: We have:
\( x^2 - 55x + 750 = 0 \)
\( \Rightarrow x^2 - 30x - 25x + 750 = 0 \) [Since \( (-30) \times (-25) = 750 \) and \( (-30) + (-25) = -55 \)]
\( \Rightarrow x(x - 30) - 25(x - 30) = 0 \)
\( \Rightarrow (x - 30)(x - 25) = 0 \)
Either \( x - 30 = 0 \Rightarrow x = 30 \)
or \( x - 25 = 0 \Rightarrow x = 25 \)
Thus, \( x = 30 \) and \( x = 25 \).

Question. Find two numbers whose sum is 27 and product is 182.
Answer: Here, sum of the numbers is 27.
Let one of the numbers be \( x \).
\( \therefore \) Other number = \( 27 - x \)
According to the condition,
Product of the numbers = 182
\( \Rightarrow x(27 - x) = 182 \)
\( \Rightarrow 27x - x^2 = 182 \)
\( \Rightarrow -x^2 + 27x - 182 = 0 \)
\( \Rightarrow x^2 - 27x + 182 = 0 \)
\( \Rightarrow x^2 - 13x - 14x + 182 = 0 \) [Since \( -27 = (-13) + (-14) \) and \( (-13) \times (-14) = 182 \)]
\( \Rightarrow x(x - 13) - 14(x - 13) = 0 \)
\( \Rightarrow (x - 13)(x - 14) = 0 \)
Either \( x - 13 = 0 \Rightarrow x = 13 \)
or \( x - 14 = 0 \Rightarrow x = 14 \)
Thus, the required numbers are 13 and 14.

Question. Find two consecutive positive integers, sum of whose squares is 365.
Answer: Let the two consecutive positive integers be \( x \) and \( (x + 1) \).
Since the sum of the squares of the numbers = 365
\( \therefore x^2 + (x + 1)^2 = 365 \)
\( \Rightarrow x^2 + [x^2 + 2x + 1] = 365 \)
\( \Rightarrow 2x^2 + 2x + 1 - 365 = 0 \)
\( \Rightarrow 2x^2 + 2x - 364 = 0 \)
\( \Rightarrow x^2 + x - 182 = 0 \)
\( \Rightarrow x^2 + 14x - 13x - 182 = 0 \) [Since \( +14 - 13 = 1 \) and \( 14 \times (-13) = -182 \)]
\( \Rightarrow x(x + 14) - 13(x + 14) = 0 \)
\( \Rightarrow (x + 14)(x - 13) = 0 \)
Either \( x + 14 = 0 \Rightarrow x = -14 \)
or \( x - 13 = 0 \Rightarrow x = 13 \)
Since \( x \) has to be a positive integer
\( \therefore x = 13 \)
\( \Rightarrow x + 1 = 13 + 1 = 14 \)
Thus, the required consecutive positive integers are 13 and 14.

Question. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Answer: Let the base of the given right triangle be ‘\( x \)’ cm.
\( \therefore \) Its height = \( (x - 7) \) cm
Given Hypotenuse \( = 13 \) cm
By Pythagoras Theorem: \( \text{Base}^2 + \text{Height}^2 = \text{Hypotenuse}^2 \)
\( \therefore x^2 + (x - 7)^2 = 13^2 \)
\( \Rightarrow x^2 + x^2 - 14x + 49 = 169 \)
\( \Rightarrow 2x^2 - 14x + 49 - 169 = 0 \)
\( \Rightarrow 2x^2 - 14x - 120 = 0 \)
\( \Rightarrow x^2 - 7x - 60 = 0 \) [Since \( (-12) \times 5 = -60 \) and \( (-12) + 5 = -7 \)]
\( \Rightarrow x^2 - 12x + 5x - 60 = 0 \)
\( \Rightarrow x(x - 12) + 5(x - 12) = 0 \)
\( \Rightarrow (x - 12)(x + 5) = 0 \)
Either \( x - 12 = 0 \Rightarrow x = 12 \)
or \( x + 5 = 0 \Rightarrow x = -5 \)
But the side of triangle can never be negative,
\( \Rightarrow x = 12 \)
\( \therefore \) Length of the base = 12 cm
\( \Rightarrow \) Length of the height = \( (12 - 7) \) cm = 5 cm
Thus, the required base = 12 cm and height = 5 cm.

Question. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.
Answer: Let the number of articles produced in a day = \( x \)
\( \therefore \) Cost of production of each article = \( Rs (2x + 3) \)
According to the condition,
Total cost = 90
\( \Rightarrow x \times (2x + 3) = 90 \)
\( \Rightarrow 2x^2 + 3x = 90 \)
\( \Rightarrow 2x^2 + 3x - 90 = 0 \)
\( \Rightarrow 2x^2 - 12x + 15x - 90 = 0 \)
\( \Rightarrow 2x(x - 6) + 15(x - 6) = 0 \)
\( \Rightarrow (x - 6)(2x + 15) = 0 \)
Either \( x - 6 = 0 \Rightarrow x = 6 \)
or \( 2x + 15 = 0 \Rightarrow x = \frac{-15}{2} \)
But the number of articles cannot be negative.
\( \therefore x = \frac{-15}{2} \) is not required
\( \Rightarrow x = 6 \)
\( \therefore \) Cost of each article = \( Rs (2 \times 6 + 3) = Rs 15 \)
Thus, the required number of articles produced is 6 and the cost of each article is Rs 15.

Solving a quadratic equation by completing the square:

Steps involved in solving \( ax^2 + bx + c = 0 \):

Divide the equation by \( a \): \( x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \)
Write \( \frac{b}{a} \) as \( 2\left(\frac{b}{2a}\right)(1) \), and \( \frac{c}{a} \) as \( -\left(\frac{b}{2a}\right)^2 + \left(\frac{b}{2a}\right)^2 + \frac{c}{a} \).
Therefore, the equation becomes: \( x^2 + 2\left(\frac{b}{2a}\right)(1) + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a} = 0 \)
Here, the first three terms are a perfect square.
\( \Rightarrow \left(x + \frac{b}{2a}\right)^2 - \left[\left(\frac{b}{2a}\right)^2 - \frac{c}{a}\right] = 0 \)
Now this can be solved easily to first find the value of \( x + \frac{b}{2a} \) and then \( x \).

Quadratic Formula:

The roots of \( ax^2 + bx + c = 0 \) are also given simply by the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
i.e., \( x = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \) or \( x = \frac{-b - \sqrt{b^2 - 4ac}}{2a} \)
The term \( b^2 - 4ac \) is said to be the discriminant \( D \).
(i) If \( D > 0 \), roots exist and they are distinct.
(ii) If \( D = 0 \), the two roots are equal and real.
(iii) If \( D < 0 \), \( \sqrt{b^2 - 4ac} \) does not exist, so there are no real roots.

Question. Find the roots of the following quadratic equation, if they exist, by the method of completing the square: \( 2x^2 - 7x + 3 = 0 \)
Answer: Sol. \( 2x^2 - 7x + 3 = 0 \)
Dividing throughout by the co-efficient of \( x^2 \), we get
\( x^2 - \frac{7}{2}x + \frac{3}{2} = 0 \)
\( \Rightarrow \left\{ x - \frac{1}{2} \left( \frac{7}{2} \right) \right\}^2 - \left( \frac{7}{4} \right)^2 + \frac{3}{2} = 0 \)
\( \Rightarrow \left( x - \frac{7}{4} \right)^2 - \frac{49}{16} + \frac{3}{2} = 0 \)
\( \Rightarrow \left( x - \frac{7}{4} \right)^2 - \frac{49}{16} + \frac{24}{16} = 0 \)
\( \Rightarrow \left( x - \frac{7}{4} \right)^2 - \frac{25}{16} = 0 \)
\( \Rightarrow \left( x - \frac{7}{4} \right)^2 = \frac{25}{16} = \left( \frac{5}{4} \right)^2 \)
\( \Rightarrow x - \frac{7}{4} = \pm \frac{5}{4} \)
Case I:
When \( \frac{5}{4} \) is \( +ve \), then
\( x - \frac{7}{4} = \frac{5}{4} \Rightarrow x = \frac{5}{4} + \frac{7}{4} \)
\( \Rightarrow x = \frac{12}{4} = 3 \)
Case II:
When \( \frac{5}{4} \) is \( -ve \), then
\( x - \frac{7}{4} = -\frac{5}{4} \Rightarrow x = -\frac{5}{4} + \frac{7}{4} \)
\( \Rightarrow x = \frac{2}{4} = \frac{1}{2} \)
Thus, required roots are \( x = 3 \) and \( x = \frac{1}{2} \).

Question. Find the roots of the following quadratic equation, if they exist, by the method of completing the square: \( 2x^2 + x - 4 = 0 \)
Answer: Sol. We have:
\( 2x^2 + x - 4 = 0 \)
Dividing throughout by 2,
\( x^2 + \frac{x}{2} - 2 = 0 \)
\( \Rightarrow \left\{ x + \frac{1}{2} \cdot \frac{1}{2} \right\}^2 - \left( \frac{1}{4} \right)^2 - 2 = 0 \)
\( \Rightarrow \left( x + \frac{1}{4} \right)^2 - \frac{1}{16} - 2 = 0 \)
\( \Rightarrow \left( x + \frac{1}{4} \right)^2 - \frac{33}{16} = 0 \)
\( \Rightarrow \left( x + \frac{1}{4} \right)^2 = \frac{33}{16} = \left( \frac{\sqrt{33}}{4} \right)^2 \)
\( \Rightarrow x + \frac{1}{4} = \pm \frac{\sqrt{33}}{4} \)
Case I:
When \( \frac{\sqrt{33}}{4} \) is positive, then
\( x + \frac{1}{4} = \frac{\sqrt{33}}{4} \Rightarrow x = \frac{\sqrt{33}}{4} - \frac{1}{4} \Rightarrow x = \frac{\sqrt{33} - 1}{4} \)
Case II:
When \( \frac{\sqrt{33}}{4} \) is negative, then
\( x + \frac{1}{4} = -\frac{\sqrt{33}}{4} \Rightarrow x = -\frac{\sqrt{33}}{4} - \frac{1}{4} \Rightarrow x = \frac{-\sqrt{33} - 1}{4} \)
Thus, the required roots are \( x = \frac{\sqrt{33} - 1}{4} \) and \( x = \frac{-\sqrt{33} - 1}{4} \).

Question. Find the roots of the following quadratic equation, if they exist, by the method of completing the square: \( 4x^2 + 4\sqrt{3}x + 3 = 0 \)
Answer: Sol. Dividing throughout by 4, we have:
\( x^2 + \sqrt{3}x + \frac{3}{4} = 0 \)
\( \Rightarrow \left[ x + \left( \frac{1}{2} \cdot \sqrt{3} \right) \right]^2 - \left( \frac{\sqrt{3}}{2} \right)^2 + \frac{3}{4} = 0 \)
\( \Rightarrow \left[ x + \frac{\sqrt{3}}{2} \right]^2 - \frac{3}{4} + \frac{3}{4} = 0 \)
\( \Rightarrow \left[ x + \frac{\sqrt{3}}{2} \right]^2 = 0 \)
\( \Rightarrow \left( x + \frac{\sqrt{3}}{2} \right) \left( x + \frac{\sqrt{3}}{2} \right) = 0 \)
\( \Rightarrow x = -\frac{\sqrt{3}}{2} \) and \( x = -\frac{\sqrt{3}}{2} \).

Question. Find the roots of the following quadratic equation, if they exist, by the method of completing the square: \( 2x^2 + x + 4 = 0 \)
Answer: Sol. Dividing throughout by 2, we have:
\( x^2 + \frac{x}{2} + 2 = 0 \)
\( \Rightarrow \left[ x + \frac{1}{4} \right]^2 - \left( \frac{1}{4} \right)^2 + 2 = 0 \)
\( \Rightarrow \left[ x + \frac{1}{4} \right]^2 - \frac{1}{16} + 2 = 0 \)
\( \Rightarrow \left[ x + \frac{1}{4} \right]^2 + \frac{31}{16} = 0 \)
\( \Rightarrow \left[ x + \frac{1}{4} \right]^2 = -\frac{31}{16} \)
But the square of a number cannot be negative.
\( \therefore \left[ x + \frac{1}{4} \right]^2 \) cannot be a real value.
So, no real roots exist.
\( \Rightarrow \) There is no real value of \( x \) satisfying the given equation.

Question. Find the roots of the following quadratic equations, using the quadratic formula: \( 2x^2 - 7x + 3 = 0 \)
Answer: Sol. Comparing the given equation with \( ax^2 + bx + c = 0 \), we have:
\( a = 2, b = -7, c = 3 \)
\( \therefore b^2 - 4ac = (-7)^2 - 4(2)(3) = 49 - 24 = 25 \geq 0 \)
Since \( b^2 - 4ac > 0 \), the given equation has real roots. The roots are given by
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \Rightarrow x = \frac{-(-7) \pm \sqrt{25}}{2(2)} = \frac{7 \pm 5}{4} \)
Taking \( +ve \) sign, \( x = \frac{7 + 5}{4} = \frac{12}{4} = 3 \)
Taking \( -ve \) sign, \( x = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2} \)
Thus, the roots of the given equation are \( x = 3 \) and \( x = \frac{1}{2} \).

Question. Find the roots of the following quadratic equations, using the quadratic formula: \( 2x^2 + x - 4 = 0 \)
Answer: Sol. Comparing the given equation with \( ax^2 + bx + c = 0 \), we have:
\( a = 2, b = 1, c = -4 \)
\( \therefore b^2 - 4ac = (1)^2 - 4(2)(-4) = 1 + 32 = 33 > 0 \)
Since \( b^2 - 4ac > 0 \), the given equation has real roots. The roots are given by
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{33}}{2(2)} = \frac{-1 \pm \sqrt{33}}{4} \)
Taking \( +ve \) sign, \( x = \frac{-1 + \sqrt{33}}{4} \)
Taking \( -ve \) sign, \( x = \frac{-1 - \sqrt{33}}{4} \)
Thus, the required roots are \( x = \frac{-1 + \sqrt{33}}{4} \) and \( x = \frac{-1 - \sqrt{33}}{4} \).

Question. Find the roots of the following quadratic equations, using the quadratic formula: \( 4x^2 + 4\sqrt{3}x + 3 = 0 \)
Answer: Sol. Comparing the given equation with \( ax^2 + bx + c = 0 \), we have:
\( a = 4, b = 4\sqrt{3}, c = 3 \)
\( \therefore b^2 - 4ac = (4\sqrt{3})^2 - 4(4)(3) = [16 \times 3] - 48 = 48 - 48 = 0 \)
Since \( b^2 - 4ac = 0 \), the given equation has real roots, which are given by:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4\sqrt{3} \pm \sqrt{0}}{2(4)} = \frac{-4\sqrt{3} \pm 0}{8} = \frac{-\sqrt{3} \pm 0}{2} \)
\( \therefore x = -\frac{\sqrt{3}}{2} \) and \( x = -\frac{\sqrt{3}}{2} \).

Question. Find the roots of the following quadratic equations, using the quadratic formula: \( 2x^2 + x + 4 = 0 \)
Answer: Sol. Comparing the given equation with \( ax^2 + bx + c = 0 \), we have:
\( a = 2, b = 1, c = 4 \)
\( \therefore b^2 - 4ac = (1)^2 - 4(2)(4) = 1 - 32 = -31 < 0 \)
Since \( b^2 - 4ac \) is less than 0, therefore the given equation does not have real roots.

Question. Find the roots of the following equation: \( x - \frac{1}{x} = 3, x \neq 0 \)
Answer: Sol. Here, we have:
\( x - \frac{1}{x} = 3 \Rightarrow x^2 - 1 = 3x \Rightarrow x^2 - 3x - 1 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \), we have:
\( a = 1, b = -3, c = -1 \)
\( \therefore b^2 - 4ac = (-3)^2 - 4(1)(-1) = 9 + 4 = 13 > 0 \)
\( \therefore x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-3) \pm \sqrt{13}}{2(1)} = \frac{3 \pm \sqrt{13}}{2} \)
Now, taking \( +ve \) sign, \( x = \frac{3 + \sqrt{13}}{2} \)
Taking \( -ve \) sign, \( x = \frac{3 - \sqrt{13}}{2} \)
Thus, the required roots of the given equation are \( x = \frac{3 + \sqrt{13}}{2} \) and \( x = \frac{3 - \sqrt{13}}{2} \).

Question. Find the roots of the following equation: \( \frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30}; x \neq -4, 7 \)
Answer: Sol. We have:
\( \frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30} \)
\( \Rightarrow \frac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \frac{11}{30} \)
\( \Rightarrow \frac{x - 7 - x - 4}{x^2 - 3x - 28} = \frac{11}{30} \)
\( \Rightarrow -11 \times 30 = 11(x^2 - 3x - 28) \)
\( \Rightarrow -30 = x^2 - 3x - 28 \)
\( \Rightarrow x^2 - 3x - 28 + 30 = 0 \)
\( \Rightarrow x^2 - 3x + 2 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \), we have:
\( a = 1, b = -3, c = 2 \)
\( \therefore b^2 - 4ac = (-3)^2 - 4(1)(2) = 9 - 8 = 1 > 0 \)
The quadratic equation (1) has real roots, which are given by:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-3) \pm \sqrt{1}}{2(1)} = \frac{3 \pm 1}{2} \)
Taking \( +ve \) sign, \( x = \frac{3 + 1}{2} = \frac{4}{2} = 2 \)
Taking \( -ve \) sign, \( x = \frac{3 - 1}{2} = \frac{2}{2} = 1 \)
Thus, the roots of the given equation are \( x = 2 \) and \( x = 1 \).

Question. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \( \frac{1}{3} \). Find his present age.
Answer: Sol. Let the present age of Rehman = \( x \)
\( \therefore \) 3 years ago Rehman’s age = \( (x - 3) \) years
5 years later Rehman’s age = \( (x + 5) \) years
Now, according to the condition,
\( \frac{1}{x - 3} + \frac{1}{x + 5} = \frac{1}{3} \)
\( \Rightarrow \frac{(x + 5) + (x - 3)}{(x - 3)(x + 5)} = \frac{1}{3} \)
\( \Rightarrow 3[x + 5 + x - 3] = (x - 3)(x + 5) \)
\( \Rightarrow 3[2x + 2] = x^2 + 2x - 15 \)
\( \Rightarrow 6x + 6 = x^2 + 2x - 15 \)
\( \Rightarrow x^2 + 2x - 6x - 15 - 6 = 0 \)
\( \Rightarrow x^2 - 4x - 21 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \), we have:
\( a = 1, b = -4, c = -21 \)
\( \therefore b^2 - 4ac = (-4)^2 - 4(1)(-21) = 16 + 84 = 100 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-4) \pm \sqrt{100}}{2(1)} = \frac{4 \pm 10}{2} \)
Taking positive sign, \( x = \frac{4 + 10}{2} = \frac{14}{2} = 7 \)
Taking negative sign, \( x = \frac{4 - 10}{2} = -\frac{6}{2} = -3 \)
Since age cannot be negative, \( \therefore x \neq -3 \Rightarrow x = 7 \)
So, the present age of Rehman = 7 years.

Question. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210? Find her marks in the two subjects.
Answer: Sol. Let, Shefali’s marks in Mathematics = \( x \)
\( \therefore \) Marks in English = \( (30 - x) \)
Now, according to the condition,
\( (x + 2) \times [(30 - x) - 3] = 210 \)
\( \Rightarrow (x + 2) \times (30 - x - 3) = 210 \)
\( \Rightarrow (x + 2)(-x + 27) = 210 \)
\( \Rightarrow -x^2 + 27x - 2x + 54 = 210 \)
\( \Rightarrow -x^2 + 25x + 54 - 210 = 0 \)
\( \Rightarrow -x^2 + 25x - 156 = 0 \)
\( \Rightarrow x^2 - 25x + 156 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \): \( a = 1, b = -25, c = 156 \)
\( \therefore b^2 - 4ac = (-25)^2 - 4(1)(156) = 625 - 624 = 1 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-25) \pm \sqrt{1}}{2(1)} = \frac{25 \pm 1}{2} \)
Taking \( +ve \) sign, \( x = \frac{25 + 1}{2} = \frac{26}{2} = 13 \)
Taking \( -ve \) sign, \( x = \frac{25 - 1}{2} = \frac{24}{2} = 12 \)
When \( x = 13 \), then English marks = \( 30 - 13 = 17 \)
When \( x = 12 \), then English marks = \( 30 - 12 = 18 \)
Thus, marks in Maths = 13, marks in English = 17 or marks in Maths = 12, marks in English = 18.

Question. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Answer: Sol. Let the shorter side (i.e., breadth) = \( x \) metres.
\( \therefore \) The longer side (length) = \( (x + 30) \) metres.
In a rectangle, \( diagonal = \sqrt{(breadth)^2 + (length)^2} \)
\( \Rightarrow x + 60 = \sqrt{x^2 + (x + 30)^2} \)
\( \Rightarrow (x + 60)^2 = x^2 + x^2 + 60x + 900 \)
\( \Rightarrow x^2 + 120x + 3600 = 2x^2 + 60x + 900 \)
\( \Rightarrow 2x^2 - x^2 + 60x - 120x + 900 - 3600 = 0 \)
\( \Rightarrow x^2 - 60x - 2700 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \): \( a = 1, b = -60, c = -2700 \)
\( \therefore b^2 - 4ac = (-60)^2 - 4(1)(-2700) = 3600 + 10800 = 14400 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-60) \pm \sqrt{14400}}{2(1)} = \frac{60 \pm 120}{2} \)
Taking \( +ve \) sign, \( x = \frac{60 + 120}{2} = \frac{180}{2} = 90 \)
Taking \( -ve \) sign, \( x = \frac{60 - 120}{2} = -\frac{60}{2} = -30 \)
Since breadth cannot be negative, \( \therefore x \neq -30 \Rightarrow x = 90 \)
\( \therefore x + 30 = 90 + 30 = 120 \)
Thus, the shorter side = 90 m and the longer side = 120 m.

Question. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Answer: Sol. Let the larger number be \( x \).
Since, \( (smaller number)^2 = 8(larger number) = 8x \)
Now, according to the condition,
\( x^2 - (smaller number)^2 = 180 \)
\( \Rightarrow x^2 - 8x = 180 \Rightarrow x^2 - 8x - 180 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \): \( a = 1, b = -8, c = -180 \)
\( \therefore b^2 - 4ac = (-8)^2 - 4(1)(-180) = 64 + 720 = 784 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-8) \pm \sqrt{784}}{2(1)} = \frac{8 \pm 28}{2} \)
Taking \( +ve \) sign, \( x = \frac{8 + 28}{2} = \frac{36}{2} = 18 \)
Taking \( -ve \) sign, \( x = \frac{8 - 28}{2} = -\frac{20}{2} = -10 \)
But \( x = -10 \) is not admissible because square of smaller number \( 8 \times (-10) = -80 \) is not possible.
\( \therefore \) The larger number = 18
\( \therefore \) Square of smaller number = \( 8 \times 18 = 144 \Rightarrow \) smaller number = \( \sqrt{144} = \pm 12 \)
Thus, the two numbers are: 18 and 12 or 18 and -12.

Question. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Answer: Sol. Let the uniform speed of the train be \( x \) km/hr.
Time taken = \( \frac{Distance}{Speed} = \frac{360}{x} \) hr
When speed is 5 km/hr more, then time is 1 hour less.
\( \Rightarrow \frac{360}{x + 5} = \frac{360}{x} - 1 \)
\( \Rightarrow \frac{360}{x} - \frac{360}{x + 5} = 1 \)
\( \Rightarrow 360 \left[ \frac{1}{x} - \frac{1}{x + 5} \right] = 1 \)
\( \Rightarrow \frac{(x + 5) - x}{x(x + 5)} = \frac{1}{360} \)
\( \Rightarrow 5 \times 360 = x^2 + 5x \)
\( \Rightarrow x^2 + 5x - 1800 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \): \( a = 1, b = 5, c = -1800 \)
\( \therefore b^2 - 4ac = (5)^2 - 4(1)(-1800) = 25 + 7200 = 7225 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{7225}}{2(1)} = \frac{-5 \pm 85}{2} \)
Taking \( +ve \) sign, \( x = \frac{-5 + 85}{2} = \frac{80}{2} = 40 \)
Taking \( -ve \) sign, \( x = \frac{-5 - 85}{2} = -45 \)
Since, speed cannot be negative, \( x = 40 \).
Thus, speed of the train = 40 km/hr.

Question. Two water taps together can fill a tank in \( 9\frac{3}{8} \) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Answer: Sol. Let the smaller tap fills the tank in \( x \) hours. \( \therefore \) The larger tap fills the tank in \( (x - 10) \) hours.
Time taken together in 1 hour = \( \frac{1}{x} + \frac{1}{x - 10} = \frac{x - 10 + x}{x(x - 10)} = \frac{2x - 10}{x^2 - 10x} \)
According to the condition, \( \frac{75}{8} \left( \frac{2x - 10}{x^2 - 10x} \right) = 1 \)
\( \Rightarrow \frac{150x - 750}{8x^2 - 80x} = 1 \Rightarrow 8x^2 - 80x = 150x - 750 \)
\( \Rightarrow 8x^2 - 230x + 750 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \): \( a = 8, b = -230, c = 750 \)
\( \therefore b^2 - 4ac = (-230)^2 - 4(8)(750) = 52900 - 24000 = 28900 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{230 \pm \sqrt{28900}}{16} = \frac{230 \pm 170}{16} \)
Taking \( +ve \) sign, \( x = \frac{400}{16} = 25 \)
Taking \( -ve \) sign, \( x = \frac{60}{16} = \frac{15}{4} \)
For \( x = \frac{15}{4} \), \( x - 10 = \frac{15}{4} - 10 = -\frac{25}{4} \), which is impossible as time cannot be negative.
\( \therefore x = 25 \Rightarrow x - 10 = 15 \)
Thus, smaller tap fills in 25 hours and larger tap in 15 hours.

Question. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Answer: Sol. Let the average speed of the passenger train be \( x \) km/hr.
\( \therefore \) Average speed of the express train = \( (x + 11) \) km/hr.
According to the condition, \( \frac{132}{x} - \frac{132}{x + 11} = 1 \)
\( \Rightarrow 132 \left[ \frac{(x + 11) - x}{x(x + 11)} \right] = 1 \)
\( \Rightarrow 132 \times 11 = x^2 + 11x \Rightarrow x^2 + 11x - 1452 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \): \( a = 1, b = 11, c = -1452 \)
\( \therefore b^2 - 4ac = (11)^2 - 4(1)(-1452) = 121 + 5808 = 5929 \)
\( x = \frac{-11 \pm \sqrt{5929}}{2} = \frac{-11 \pm 77}{2} \)
Taking \( +ve \) sign, \( x = \frac{66}{2} = 33 \)
Taking \( -ve \) sign, \( x = -\frac{88}{2} = -44 \)
Since speed cannot be negative, \( x = 33 \).
Passenger train speed = 33 km/hr, Express train speed = \( 33 + 11 = 44 \) km/hr.

Question. Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Answer: Sol. Let the side of smaller square be \( x \) m. Perimeter = \( 4x \) m.
Perimeter of larger square = \( (4x + 24) \) m. Side of larger square = \( \frac{4x + 24}{4} = (x + 6) \) m.
According to condition, \( x^2 + (x + 6)^2 = 468 \)
\( \Rightarrow x^2 + x^2 + 12x + 36 = 468 \Rightarrow 2x^2 + 12x - 432 = 0 \)
\( \Rightarrow x^2 + 6x - 216 = 0 \) ...(1)
Comparing (1) with \( ax^2 + bx + c = 0 \): \( a = 1, b = 6, c = -216 \)
\( \therefore b^2 - 4ac = (6)^2 - 4(1)(-216) = 36 + 864 = 900 \)
\( x = \frac{-6 \pm \sqrt{900}}{2} = \frac{-6 \pm 30}{2} \)
Taking \( +ve \) sign, \( x = \frac{24}{2} = 12 \)
Taking \( -ve \) sign, \( x = -\frac{36}{2} = -18 \)
Length cannot be negative, \( \therefore x = 12 \).
Smaller square side = 12 m, larger square side = \( 12 + 6 = 18 \) m.

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Mathematics Class 10 Exam Resources: Chapter 04 Quadratic Equations

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