CBSE Class 10 Mathematics Quadratic Equations Assignment Set 26

Exercise 1: Subjective Questions

Question 1. Which of the following are quadratic equations? (i) \( x - \dfrac{3}{x} = x^2 \)    (ii) \( (2x + 1)(3x + 2) = 6(x - 1)(x - 2) \)
Answer:
(i) Multiplying by x: \( x^2 - 3 = x^3 \), that is \( x^3 - x^2 + 3 = 0 \), a cubic. Not a quadratic equation.
(ii) \( 6x^2 + 7x + 2 = 6x^2 - 18x + 12 \) gives \( 25x - 10 = 0 \), a linear equation. Not a quadratic equation.
In simple words: Simplify fully and check the highest power of x.

Teacher's Note:
1. The x² terms may cancel.
2. Always compare the degree after simplifying.

Question 2. Determine whether the given values are solutions of the given equation or not. (i) \( 2x^2 - 5x - 3 = 0 \); x = 3    (ii) \( x + \dfrac{1}{x} = \dfrac{13}{6} \); \( x = \dfrac{5}{6}, \ x = \dfrac{4}{3} \)
Answer:
(i) \( 2(9) - 15 - 3 = 0 \). Yes, x = 3 is a solution.
(ii) For \( x = \dfrac{5}{6} \): \( \dfrac{5}{6} + \dfrac{6}{5} = \dfrac{61}{30} \ne \dfrac{65}{30} \). For \( x = \dfrac{4}{3} \): \( \dfrac{4}{3} + \dfrac{3}{4} = \dfrac{25}{12} \ne \dfrac{13}{6} \). No, neither is a solution (the solutions are \( \dfrac{3}{2} \) and \( \dfrac{2}{3} \)).
In simple words: Substitute and compare both sides.

Teacher's Note:
1. A value is a solution only if it makes both sides equal.

Question 3. Find the value of K for which the given value is a solution of the equation. (i) \( 3x^2 + 2x - K = 0 \), \( x = \dfrac{1}{3} \)    (ii) \( Kx^2 - 3abx + 2b^2 = 0 \), \( x = \dfrac{b}{a} \)
Answer:
(i) \( 3 \cdot \dfrac{1}{9} + \dfrac{2}{3} - K = 0 \Rightarrow K = 1 \).
(ii) \( K\dfrac{b^2}{a^2} - 3ab \cdot \dfrac{b}{a} + 2b^2 = 0 \Rightarrow K\dfrac{b^2}{a^2} = b^2 \Rightarrow K = a^2 \).
In simple words: Put the given value of x into the equation.

Teacher's Note:
1. Then solve the resulting equation for K.

Question 4. Is x = -4 a solution of the equation \( 2x^2 + 5x - 12 = 0 \)?
Answer:
\( 2(16) + 5(-4) - 12 = 32 - 20 - 12 = 0 \). Yes.
In simple words: Substitute x = -4.

Teacher's Note:
1. The left side becomes 0, so -4 is a root.

Question 5. If x = -2 and \( x = \dfrac{1}{5} \) are solutions of \( 5x^2 + kx + \lambda = 0 \), find k and \( \lambda \).
Answer:
Sum of the roots \( = -2 + \dfrac{1}{5} = -\dfrac{9}{5} = -\dfrac{k}{5} \Rightarrow k = 9 \). Product \( = -\dfrac{2}{5} = \dfrac{\lambda}{5} \Rightarrow \lambda = -2 \).
k = 9, \( \lambda \) = -2.
In simple words: Use the sum and product of the roots.

Teacher's Note:
1. Check with x = -2: 20 - 18 - 2 = 0.

Question 6. Solve by the factorisation method: (i) \( \left(x - \dfrac{1}{2}\right)^2 = 4 \)    (ii) \( x^2 - 2bx + b^2 - a^2 = 0 \)
Answer:
(i) \( \left(x - \dfrac{1}{2}\right)^2 - 2^2 = 0 \Rightarrow \left(x - \dfrac{1}{2} - 2\right)\left(x - \dfrac{1}{2} + 2\right) = 0 \): \( x = \dfrac{5}{2}, -\dfrac{3}{2} \).
(ii) \( (x - b)^2 - a^2 = (x - b - a)(x - b + a) = 0 \): x = b + a or x = b - a.
In simple words: Use a² - b² = (a - b)(a + b).

Teacher's Note:
1. Both parts are differences of squares.

Question 7. Find the roots by completing the square: (i) \( 8x^2 + 2x - 3 = 0 \)    (ii) \( x^2 - 4ax + 4a^2 - b^2 = 0 \)
Answer:
(i) \( x^2 + \dfrac{x}{4} = \dfrac{3}{8} \Rightarrow \left(x + \dfrac{1}{8}\right)^2 = \dfrac{3}{8} + \dfrac{1}{64} = \dfrac{25}{64} \Rightarrow x + \dfrac{1}{8} = \pm\dfrac{5}{8} \): \( x = \dfrac{1}{2}, -\dfrac{3}{4} \).
(ii) \( (x - 2a)^2 = b^2 \Rightarrow x - 2a = \pm b \): x = 2a + b, x = 2a - b.
In simple words: Make the leading coefficient 1, then add the square of half the x-coefficient.

Teacher's Note:
1. Check (i) by substitution: 8/4 + 1 - 3 = 0.

Question 8. Find the roots by the quadratic formula: (i) \( x^2 + 2\sqrt{2}x - 6 = 0 \)    (ii) \( 5x^2 - 5x - 3 = 0 \)
Answer:
(i) \( D = 8 + 24 = 32 \), \( x = \dfrac{-2\sqrt{2} \pm 4\sqrt{2}}{2} \): \( x = \sqrt{2}, -3\sqrt{2} \).
(ii) \( D = 25 + 60 = 85 \): \( x = \dfrac{5 \pm \sqrt{85}}{10} \).
In simple words: x = (-b ± √D) / 2a.

Teacher's Note:
1. \( \sqrt{32} = 4\sqrt{2} \), so the roots in (i) simplify.

Question 9. The product of the roots of the equation \( x^2 - 8x + k = 0 \) is 12. Find the value of k.
Answer:
Product of roots \( = \dfrac{k}{1} = 12 \), so k = 12. (Check: \( x^2 - 8x + 12 = (x - 2)(x - 6) \), roots 2 and 6.)
In simple words: Product of the roots = c/a.

Teacher's Note:
1. Here a = 1, so the product equals k.

Question 10. Show that the condition for one root of the quadratic equation \( x^2 + px + q = 0 \) to be twice the other is \( 2p^2 = 9q \).
Answer:
Let the roots be \( \alpha \) and \( 2\alpha \). Sum: \( 3\alpha = -p \Rightarrow \alpha = -\dfrac{p}{3} \). Product: \( 2\alpha^2 = q \).
So \( 2 \cdot \dfrac{p^2}{9} = q \), that is \( 2p^2 = 9q \).
In simple words: Name the roots α and 2α.

Teacher's Note:
1. Eliminate α between the sum and the product.

Question 11. Find the value of k such that the equation \( x^2 + (k - 1)x - 1 = 0 \) has two roots that are equal in magnitude but opposite in sign.
Answer:
The sum of the roots must be zero: \( -(k - 1) = 0 \), so k = 1. The equation is then \( x^2 - 1 = 0 \) with roots 1 and -1.
In simple words: Opposite roots means sum = 0.

Teacher's Note:
1. Always check that the roots are real after finding k.

Question 12. The roots of \( x^2 + Ax + B = 0 \) are 5 and 4. The roots of \( x^2 + Cx + D = 0 \) are 2 and 9. Find the roots of \( x^2 + Ax + D = 0 \).
Answer:
\( A = -(5 + 4) = -9 \) and \( D = 2 \times 9 = 18 \). The equation is \( x^2 - 9x + 18 = 0 \Rightarrow (x - 3)(x - 6) = 0 \).
The roots are 6 and 3.
In simple words: Find A and D from the sum and product.

Teacher's Note:
1. Sum = -A and product = D here.

Question 13. If the roots of \( (a^2 + b^2)x^2 - 2b(a + c)x + (b^2 + c^2) = 0 \) are equal, show that \( b^2 = ac \).
Answer:
\( D = 4b^2(a + c)^2 - 4(a^2 + b^2)(b^2 + c^2) = 0 \). Divide by 4 and expand: \( b^2a^2 + 2ab^2c + b^2c^2 - a^2b^2 - a^2c^2 - b^4 - b^2c^2 = 0 \).
\( 2ab^2c - a^2c^2 - b^4 = 0 \Rightarrow -(b^2 - ac)^2 = 0 \Rightarrow b^2 = ac \).
In simple words: Put D = 0 and expand carefully.

Teacher's Note:
1. The result collapses to a perfect square.

Question 14. Find the discriminant of \( 3\sqrt{3}x^2 + 10x + \sqrt{3} = 0 \).
Answer:
\( D = 10^2 - 4(3\sqrt{3})(\sqrt{3}) = 100 - 36 = 64 \). D = 64.
In simple words: D = b² - 4ac.

Teacher's Note:
1. \( \sqrt{3} \times \sqrt{3} = 3 \), so 4 x 3 x 3 = 36.

Question 15. For what value of K does \( 9x^2 + 3Kx + 4 = 0 \) have equal roots?
Answer:
\( D = 9K^2 - 144 = 0 \Rightarrow K^2 = 16 \). K = ±4.
In simple words: Set D = 0.

Teacher's Note:
1. Equal roots need b² = 4ac.

Question 16. Find the value of k for which \( (k - 2)x^2 + 2(2k - 3)x + 5k - 6 = 0 \) has equal roots.
Answer:
\( D/4 = (2k - 3)^2 - (k - 2)(5k - 6) = 4k^2 - 12k + 9 - 5k^2 + 16k - 12 = -k^2 + 4k - 3 = 0 \Rightarrow (k - 1)(k - 3) = 0 \).
k = 1 or k = 3. (For both values \( k e 2 \), so the equation stays quadratic.)
In simple words: Use D = 0 with b = 2(2k - 3).

Teacher's Note:
1. Check that the x² coefficient is not zero.

Question 17. Show that the roots of \( x^2 + ax - 4 = 0 \) (where \( a \in \mathbb{R} \)) are real and distinct.
Answer:
\( D = a^2 + 16 \). Since \( a^2 \ge 0 \), \( D \ge 16 > 0 \) for every real a. Hence the roots are real and distinct.
In simple words: Show that D is always positive.

Teacher's Note:
1. A square plus a positive number is positive.

Question 18. A man is five times as old as his son and the sum of the squares of their ages is 2106. Find their ages.
Answer:
Son's age x, man's age 5x: \( x^2 + 25x^2 = 2106 \Rightarrow x^2 = 81 \Rightarrow x = 9 \) (age cannot be negative).
Son 9 years, man 45 years.
In simple words: Form 26x² = 2106.

Teacher's Note:
1. Reject the negative root.

Question 19. The lengths of the sides of a right triangle are 5x + 2, 5x and 3x - 1. If x > 0, find the length of each side.
Answer:
The hypotenuse is the longest side, 5x + 2: \( (5x)^2 + (3x - 1)^2 = (5x + 2)^2 \Rightarrow 9x^2 - 26x - 3 = 0 \Rightarrow (x - 3)(9x + 1) = 0 \). Since x > 0, x = 3.
The sides are 17, 15 and 8 (17² = 15² + 8²).
In simple words: Choose the hypotenuse first.

Teacher's Note:
1. Check: 289 = 225 + 64.

Question 20. The sum of two numbers is 8. Determine the numbers if the sum of their reciprocals is \( \dfrac{8}{15} \).
Answer:
Numbers x and 8 - x: \( \dfrac{1}{x} + \dfrac{1}{8 - x} = \dfrac{8}{15} \Rightarrow 15 \cdot 8 = 8x(8 - x) \Rightarrow x^2 - 8x + 15 = 0 \Rightarrow (x - 3)(x - 5) = 0 \).
The numbers are 3 and 5.
In simple words: Combine the fractions: 8/(x(8 - x)).

Teacher's Note:
1. The two roots are the two numbers.

Question 21. The numerator of a fraction is 1 less than its denominator. If 3 is added to each of the numerator and denominator, the fraction is increased by \( \dfrac{3}{28} \). Find the fraction.
Answer:
Let the denominator be d: fraction \( \dfrac{d - 1}{d} \). \( \dfrac{d + 2}{d + 3} - \dfrac{d - 1}{d} = \dfrac{3}{28} \Rightarrow \dfrac{3}{d(d + 3)} = \dfrac{3}{28} \Rightarrow d^2 + 3d - 28 = 0 \Rightarrow (d + 7)(d - 4) = 0 \).
d = 4, so the fraction is \( \dfrac{3}{4} \).
In simple words: The numerator difference simplifies to 3.

Teacher's Note:
1. Check: 5/7 - 3/4 = 3/28.

Question 22. A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Answer:
Let the stream speed be x km/h: \( \dfrac{24}{18 - x} - \dfrac{24}{18 + x} = 1 \Rightarrow 48x = 324 - x^2 \Rightarrow x^2 + 48x - 324 = 0 \Rightarrow (x + 54)(x - 6) = 0 \).
The speed of the stream is 6 km/h.
In simple words: Time = distance / speed.

Teacher's Note:
1. Reject x = -54.