CBSE Class 10 Mathematics Quadratic Equations Assignment Set 22

Exercise III: Completing the Square, Quadratic Formula and Word Problems

Question 1. Find the roots of the following quadratic equation, if they exist, by the method of completing the square: \( 2x^2 - 7x + 3 = 0 \).
Answer:
Dividing by 2: \( x^2 - \frac{7}{2}x + \frac{3}{2} = 0 \Rightarrow x^2 - 2 \times x \times \frac{7}{4} = -\frac{3}{2} \).
Adding \( \left(\frac{7}{4}\right)^2 \) to both sides: \( \left(x - \frac{7}{4}\right)^2 = \frac{49}{16} - \frac{3}{2} = \frac{25}{16} \).
\( x - \frac{7}{4} = \pm\frac{5}{4} \Rightarrow x = \frac{7 + 5}{4} \) or \( \frac{7 - 5}{4} \).
Hence \( x = 3 \) or \( x = \frac{1}{2} \).
In simple words: Make the coefficient of x² equal to 1 and complete the square.

Teacher's Note:
1. Write the equation as x² + (b/a)x = −c/a, add the square of half the coefficient of x to both sides and take square roots.
2. Check by substituting the roots in the original equation.

Question 2. Find the roots of the following quadratic equation, if they exist, by the method of completing the square: \( 2x^2 + x - 4 = 0 \).
Answer:
Dividing by 2: \( x^2 + \frac{1}{2}x = 2 \).
Adding \( \left(\frac{1}{4}\right)^2 \) to both sides: \( \left(x + \frac{1}{4}\right)^2 = 2 + \frac{1}{16} = \frac{33}{16} \).
\( x + \frac{1}{4} = \pm\frac{\sqrt{33}}{4} \Rightarrow x = \frac{-1 \pm \sqrt{33}}{4} \).
Hence \( x = \frac{-1 + \sqrt{33}}{4} \) or \( x = \frac{-1 - \sqrt{33}}{4} \).
In simple words: The right side becomes 33/16, which is not a perfect square, so the roots are irrational.

Teacher's Note:
1. Write the equation as x² + (b/a)x = −c/a, add the square of half the coefficient of x to both sides and take square roots.
2. Check by substituting the roots in the original equation.

Question 3. 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:
Dividing by 4: \( x^2 + \sqrt{3}x = -\frac{3}{4} \).
Adding \( \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \) to both sides: \( \left(x + \frac{\sqrt{3}}{2}\right)^2 = -\frac{3}{4} + \frac{3}{4} = 0 \).
Hence \( x = -\frac{\sqrt{3}}{2} \) (twice).
In simple words: The right side becomes 0, so both roots are equal.

Teacher's Note:
1. Write the equation as x² + (b/a)x = −c/a, add the square of half the coefficient of x to both sides and take square roots.
2. Check by substituting the roots in the original equation.

Question 4. Find the roots of the following quadratic equation, if they exist, by the method of completing the square: \( 2x^2 + x + 4 = 0 \).
Answer:
Dividing by 2: \( x^2 + \frac{1}{2}x = -2 \).
Adding \( \left(\frac{1}{4}\right)^2 \): \( \left(x + \frac{1}{4}\right)^2 = -2 + \frac{1}{16} = -\frac{31}{16} \).
The square of a real number cannot be negative, so no real roots exist.
In simple words: A square equal to a negative number is impossible, so there are no real roots.

Teacher's Note:
1. Write the equation as x² + (b/a)x = −c/a, add the square of half the coefficient of x to both sides and take square roots.
2. Check by substituting the roots in the original equation.

Question 5. Find the roots of the quadratic equation (from Q.1) by applying the quadratic formula: \( 2 x^{2} - 7 x + 3 = 0 \).
Answer:
Comparing with \( ax^2 + bx + c = 0 \): \( a = 2,\ b = -7,\ c = 3 \).
\( D = b^2 - 4ac = (-7)^2 - 4(2)(3) = 25 \).
\( x = \frac{-b \pm \sqrt{D}}{2a} = \frac{7 \pm 5}{4} \).
Hence \( x = \frac{1}{2}\ \text{or}\ x = 3 \).
In simple words: Substitute a, b and c in the quadratic formula.

Teacher's Note:
1. Find a, b, c, then D = b² − 4ac; if D ≥ 0 use x = (−b ± √D)/2a.
2. If D < 0 write that no real roots exist.

Question 6. Find the roots of the quadratic equation (from Q.1) by applying the quadratic formula: \( 2 x^{2} + x - 4 = 0 \).
Answer:
Comparing with \( ax^2 + bx + c = 0 \): \( a = 2,\ b = 1,\ c = -4 \).
\( D = b^2 - 4ac = (1)^2 - 4(2)(-4) = 33 \).
\( x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-1 \pm \sqrt{33}}{4} \).
Hence \( x = - \frac{\sqrt{33}}{4} - \frac{1}{4}\ \text{or}\ x = - \frac{1}{4} + \frac{\sqrt{33}}{4} \).
In simple words: Substitute a, b and c in the quadratic formula.

Teacher's Note:
1. Find a, b, c, then D = b² − 4ac; if D ≥ 0 use x = (−b ± √D)/2a.
2. If D < 0 write that no real roots exist.

Question 7. Find the roots of the quadratic equation (from Q.1) by applying the quadratic formula: \( 4 x^{2} + 4 \sqrt{3} x + 3 = 0 \).
Answer:
Comparing with \( ax^2 + bx + c = 0 \): \( a = 4,\ b = 4 \sqrt{3},\ c = 3 \).
\( D = b^2 - 4ac = (4 \sqrt{3})^2 - 4(4)(3) = 0 \).
\( x = \frac{-b \pm \sqrt{D}}{2a} = \frac{- 4 \sqrt{3} \pm 0}{8} \).
Hence \( x = - \frac{\sqrt{3}}{2} \).
In simple words: Substitute a, b and c in the quadratic formula.

Teacher's Note:
1. Find a, b, c, then D = b² − 4ac; if D ≥ 0 use x = (−b ± √D)/2a.
2. If D < 0 write that no real roots exist.

Question 8. Find the roots of the quadratic equation (from Q.1) by applying the quadratic formula: \( 2 x^{2} + x + 4 = 0 \).
Answer:
Comparing with \( ax^2 + bx + c = 0 \): \( a = 2,\ b = 1,\ c = 4 \).
\( D = b^2 - 4ac = (1)^2 - 4(2)(4) = -31 \).
Since \( D < 0 \), the equation has no real roots.
In simple words: Substitute a, b and c in the quadratic formula.

Teacher's Note:
1. Find a, b, c, then D = b² − 4ac; if D ≥ 0 use x = (−b ± √D)/2a.
2. If D < 0 write that no real roots exist.

Question 9. Find the roots of the following equation: \( x - \frac{1}{x} = 3,\ x \neq 0 \).
Answer:
Multiplying by x: \( x^2 - 1 = 3x \Rightarrow x^2 - 3x - 1 = 0 \).
Here \( a = 1,\ b = -3,\ c = -1 \). \( D = 9 + 4 = 13 \).
\( x = \frac{3 \pm \sqrt{13}}{2} \).
Hence \( x = \frac{3 + \sqrt{13}}{2} \) or \( x = \frac{3 - \sqrt{13}}{2} \).
In simple words: Multiply through by x to get a quadratic equation and use the formula.

Teacher's Note:
1. Multiply by the common denominator to clear fractions.
2. Remember the condition x ≠ 0 given in the question.

Question 10. Find the roots of the following equation: \( \frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30},\ x \neq -4,\ 7 \).
Answer:
\( \frac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \frac{11}{30} \Rightarrow \frac{-11}{(x + 4)(x - 7)} = \frac{11}{30} \).
\( (x + 4)(x - 7) = -30 \Rightarrow x^2 - 3x - 28 = -30 \Rightarrow x^2 - 3x + 2 = 0 \).
\( (x - 2)(x - 1) = 0 \Rightarrow x = 1 \) or \( x = 2 \).
Hence \( x = 1 \) or \( x = 2 \). Both are allowed as neither is \( -4 \) or 7.
In simple words: Combine the fractions on the left, cross-multiply and solve the quadratic.

Teacher's Note:
1. Cross-multiplying removes the denominators.
2. Check the roots against the excluded values.

Question 11. Solve the following problem: 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:
Let Rehman's present age be x years. Three years ago his age was \( (x - 3) \) years and five years from now it will be \( (x + 5) \) years.
\( \frac{1}{x - 3} + \frac{1}{x + 5} = \frac{1}{3} \Rightarrow \frac{2x + 2}{(x - 3)(x + 5)} = \frac{1}{3} \Rightarrow 3(2x + 2) = x^2 + 2x - 15 \).
\( x^2 - 4x - 21 = 0 \Rightarrow x^2 - 7x + 3x - 21 = 0 \Rightarrow (x - 7)(x + 3) = 0 \Rightarrow x = 7 \) or \( x = -3 \).
Age cannot be negative, so Rehman's present age is 7 years. Check: \( \frac{1}{4} + \frac{1}{12} = \frac{1}{3} \).
In simple words: Ages 3 years ago and 5 years later are x - 3 and x + 5.

Teacher's Note:
1. Form the quadratic equation from the given condition, solve it and reject any root that is not possible for the situation.
2. Verify the answer in the original statement.

Question 12. Solve the following problem: 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 her marks would have been 210. Find her marks in the two subjects.
Answer:
Let her marks in Mathematics be x. Then her marks in English are \( 30 - x \).
Changed marks: \( (x + 2) \) and \( (30 - x - 3) = (27 - x) \). Product \( = 210 \Rightarrow (x + 2)(27 - x) = 210 \).
\( 27x - x^2 + 54 - 2x = 210 \Rightarrow x^2 - 25x + 156 = 0 \Rightarrow x^2 - 12x - 13x + 156 = 0 \Rightarrow (x - 12)(x - 13) = 0 \).
So \( x = 12 \) or \( x = 13 \).
If \( x = 12 \): Mathematics 12 and English 18. If \( x = 13 \): Mathematics 13 and English 17.
Hence her marks are 12 in Mathematics and 18 in English, or 13 in Mathematics and 17 in English.
In simple words: Her new marks are x + 2 and 27 - x; both answers fit the conditions.

Teacher's Note:
1. Form the quadratic equation from the given condition, solve it and reject any root that is not possible for the situation.
2. Verify the answer in the original statement.
3. Both roots are positive and less than 30, so both pairs are acceptable.

Question 13. Solve the following problem: 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:
Let the shorter side be x m. Then the longer side is \( (x + 30) \) m and the diagonal is \( (x + 60) \) m.
By the Pythagoras theorem: \( x^2 + (x + 30)^2 = (x + 60)^2 \Rightarrow 2x^2 + 60x + 900 = x^2 + 120x + 3600 \Rightarrow x^2 - 60x - 2700 = 0 \).
\( x^2 - 90x + 30x - 2700 = 0 \Rightarrow (x - 90)(x + 30) = 0 \Rightarrow x = 90 \) or \( x = -30 \).
A side cannot be negative, so the shorter side is 90 m and the longer side is \( 90 + 30 = 120 \) m.
Hence the sides are 90 m and 120 m.
In simple words: Apply the Pythagoras theorem to the sides x, x + 30 and the diagonal x + 60.

Teacher's Note:
1. Reject the negative root.
2. Check: 90² + 120² = 22500 = 150².

Question 14. Solve the following problem: 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:
Let the larger number be x and the smaller number be y. Then \( x^2 - y^2 = 180 \) ...(i) and \( y^2 = 8x \) ...(ii).
Putting (ii) in (i): \( x^2 - 8x = 180 \Rightarrow x^2 - 8x - 180 = 0 \Rightarrow x^2 - 18x + 10x - 180 = 0 \Rightarrow (x - 18)(x + 10) = 0 \).
\( x = 18 \) or \( x = -10 \). For \( x = -10 \), \( y^2 = -80 \) is impossible, so \( x = 18 \).
Then \( y^2 = 8 \times 18 = 144 \Rightarrow y = \pm 12 \).
Hence the numbers are 18 and 12 (taking positive numbers; the pair 18 and \( -12 \) also satisfies both conditions).
In simple words: Use the two conditions to get an equation in the larger number alone.

Teacher's Note:
1. Substitute y² = 8x to remove y.
2. Reject the root that makes y² negative.

Question 15. Solve the following problem: 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:
Let the speed of the train be x km/h. Time taken \( = \frac{360}{x} \) hours. With speed \( (x + 5) \) km/h, time \( = \frac{360}{x + 5} \) hours.
\( \frac{360}{x} - \frac{360}{x + 5} = 1 \Rightarrow 360 \times 5 = x(x + 5) \Rightarrow x^2 + 5x - 1800 = 0 \).
\( x^2 + 45x - 40x - 1800 = 0 \Rightarrow (x + 45)(x - 40) = 0 \Rightarrow x = 40 \) (speed cannot be negative).
Hence the speed of the train is 40 km/h. Check: 9 h - 8 h = 1 h.
In simple words: Time difference is 1 hour, so 360/x - 360/(x + 5) = 1.

Teacher's Note:
1. Form the quadratic equation from the given condition, solve it and reject any root that is not possible for the situation.
2. Verify the answer in the original statement.
3. Use time = distance ÷ speed.

Question 16. Solve the following problem: 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:
Let the smaller tap take x hours. Then the larger tap takes \( (x - 10) \) hours. In 1 hour they fill \( \frac{1}{x} \) and \( \frac{1}{x - 10} \) of the tank.
Together they fill the tank in \( 9\frac{3}{8} = \frac{75}{8} \) hours, so in 1 hour they fill \( \frac{8}{75} \): \( \frac{1}{x} + \frac{1}{x - 10} = \frac{8}{75} \Rightarrow \frac{2x - 10}{x(x - 10)} = \frac{8}{75} \).
\( 75(2x - 10) = 8x^2 - 80x \Rightarrow 8x^2 - 230x + 750 = 0 \Rightarrow 4x^2 - 115x + 375 = 0 \).
\( 4x^2 - 100x - 15x + 375 = 0 \Rightarrow (x - 25)(4x - 15) = 0 \Rightarrow x = 25 \) or \( x = \frac{15}{4} \).
If \( x = \frac{15}{4} \) then \( x - 10 \) would be negative, which is impossible. So \( x = 25 \).
Hence the smaller tap takes 25 hours and the larger tap takes 15 hours.
In simple words: Add the part of the tank each tap fills in one hour and equate it to 8/75.

Teacher's Note:
1. Form the quadratic equation from the given condition, solve it and reject any root that is not possible for the situation.
2. Verify the answer in the original statement.
3. Work done in 1 hour is the reciprocal of the time taken.

Question 17. Solve the following problem: 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:
Let the average speed of the passenger train be x km/h. Then the express train's average speed is \( (x + 11) \) km/h.
\( \frac{132}{x} - \frac{132}{x + 11} = 1 \Rightarrow 132 \times 11 = x(x + 11) \Rightarrow x^2 + 11x - 1452 = 0 \).
\( x^2 + 44x - 33x - 1452 = 0 \Rightarrow (x + 44)(x - 33) = 0 \Rightarrow x = 33 \) (negative speed rejected).
Hence the passenger train's speed is 33 km/h and the express train's speed is 44 km/h.
In simple words: The slower train takes one hour longer for the same 132 km.

Teacher's Note:
1. Form the quadratic equation from the given condition, solve it and reject any root that is not possible for the situation.
2. Verify the answer in the original statement.

Question 18. Solve the following problem: The 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:
Let the side of the smaller square be x m and that of the larger square be y m. Perimeters are 4x and 4y, areas are \( x^2 \) and \( y^2 \).
\( 4y - 4x = 24 \Rightarrow y = x + 6 \). Also \( x^2 + y^2 = 468 \Rightarrow x^2 + (x + 6)^2 = 468 \Rightarrow 2x^2 + 12x - 432 = 0 \Rightarrow x^2 + 6x - 216 = 0 \).
\( x^2 + 18x - 12x - 216 = 0 \Rightarrow (x + 18)(x - 12) = 0 \Rightarrow x = 12 \) (a side cannot be negative).
Then \( y = 12 + 6 = 18 \).
Hence the sides of the squares are 12 m and 18 m. Check: 144 + 324 = 468.
In simple words: Use the perimeter difference to write one side in terms of the other and then the sum of areas.

Teacher's Note:
1. Form the quadratic equation from the given condition, solve it and reject any root that is not possible for the situation.
2. Verify the answer in the original statement.

Exercise IV: Nature of Roots

Question 19. Find the nature of the roots of the following quadratic equation. If the real roots exist, find them: \( 2x^2 - 3x + 5 = 0 \).
Answer:
Comparing with \( ax^2 + bx + c = 0 \): \( a = 2,\ b = -3,\ c = 5 \).
\( D = b^2 - 4ac = (-3)^2 - 4(2)(5) = -31 \).
Since \( D < 0 \), the equation has no real roots.
In simple words: A negative discriminant means no real roots.

Teacher's Note:
1. Compare with ax² + bx + c = 0 and find D = b² − 4ac.
2. D > 0: two distinct real roots; D = 0: two equal real roots; D < 0: no real roots.

Question 20. Find the nature of the roots of the following quadratic equation. If the real roots exist, find them: \( 3x^2 - 4\sqrt{3}x + 4 = 0 \).
Answer:
Comparing with \( ax^2 + bx + c = 0 \): \( a = 3,\ b = -4\sqrt{3},\ c = 4 \).
\( D = b^2 - 4ac = (-4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0 \).
Since \( D = 0 \), the roots are real and equal: \( x = \frac{-b}{2a} = \frac{4\sqrt{3}}{6} = \frac{2}{\sqrt{3}} \).
Hence the roots are \( \frac{2}{\sqrt{3}} \) and \( \frac{2}{\sqrt{3}} \).
In simple words: D = 0 gives two equal real roots.

Teacher's Note:
1. Compare with ax² + bx + c = 0 and find D = b² − 4ac.
2. D > 0: two distinct real roots; D = 0: two equal real roots; D < 0: no real roots.

Question 21. Find the nature of the roots of the following quadratic equation. If the real roots exist, find them: \( 2x^2 - 6x + 3 = 0 \).
Answer:
Comparing with \( ax^2 + bx + c = 0 \): \( a = 2,\ b = -6,\ c = 3 \).
\( D = (-6)^2 - 4(2)(3) = 36 - 24 = 12 > 0 \), so the roots are real and distinct.
\( x = \frac{6 \pm \sqrt{12}}{4} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2} \).
Hence the roots are \( \frac{3 + \sqrt{3}}{2} \) and \( \frac{3 - \sqrt{3}}{2} \).
In simple words: D = 12 is positive but not a perfect square, so the roots are real, distinct and irrational.

Teacher's Note:
1. Compare with ax² + bx + c = 0 and find D = b² − 4ac.
2. D > 0: two distinct real roots; D = 0: two equal real roots; D < 0: no real roots.

Question 22. Find the values of k for which the following quadratic equation has two equal roots: \( 2x^2 + kx + 3 = 0 \).
Answer:
Here \( a = 2,\ b = k,\ c = 3 \). \( D = k^2 - 4(2)(3) = k^2 - 24 \).
For equal roots, \( D = 0 \Rightarrow k^2 = 24 \Rightarrow k = \pm\sqrt{24} = \pm 2\sqrt{6} \).
Hence \( k = 2\sqrt{6} \) or \( k = -2\sqrt{6} \).
In simple words: Put the discriminant equal to zero and solve for k.

Teacher's Note:
1. Equal roots mean D = 0; form D in terms of k and solve.
2. Reject a value of k that destroys the quadratic nature of the equation.

Question 23. Find the values of k for which the following quadratic equation has two equal roots: \( kx(x - 2) + 6 = 0 \).
Answer:
The equation is \( kx^2 - 2kx + 6 = 0 \), so \( a = k,\ b = -2k,\ c = 6 \).
\( D = (-2k)^2 - 4(k)(6) = 4k^2 - 24k \). For equal roots, \( D = 0 \Rightarrow 4k(k - 6) = 0 \Rightarrow k = 0 \) or \( k = 6 \).
If \( k = 0 \), the equation becomes \( 6 = 0 \), which has no terms in \( x^2 \) and \( x \), so it is not a quadratic equation.
Hence \( k = 6 \) only.
In simple words: Expand, find D, and discard k = 0 because then the equation is no longer quadratic.

Teacher's Note:
1. Equal roots mean D = 0; form D in terms of k and solve.
2. Reject a value of k that destroys the quadratic nature of the equation.

Question 24. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.
Answer:
Let the breadth be x m. Then the length is 2x m, and area \( = 2x \times x = 800 \Rightarrow 2x^2 - 800 = 0 \Rightarrow x^2 - 400 = 0 \).
\( D = 0 - 4(1)(-400) = 1600 > 0 \), so real roots exist: \( x = \pm 20 \). A breadth cannot be negative, so \( x = 20 \).
Yes, it is possible: the breadth is 20 m and the length is 40 m.
In simple words: Form 2x squared = 800 and check that it has a positive real root.

Teacher's Note:
1. Check D ≥ 0 before claiming the situation is possible.
2. Reject the negative root for a length.

Question 25. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Answer:
Let the present age of the first friend be x years. Then the second friend's age is \( (20 - x) \) years.
Four years ago: \( (x - 4)(16 - x) = 48 \Rightarrow 16x - x^2 - 64 + 4x = 48 \Rightarrow x^2 - 20x + 112 = 0 \).
\( D = (-20)^2 - 4(1)(112) = 400 - 448 = -48 < 0 \), so the equation has no real roots.
Hence the situation is not possible.
In simple words: The quadratic formed has a negative discriminant, so no such ages exist.

Teacher's Note:
1. If D < 0 the situation described cannot happen.
2. Always find D before trying to solve.

Question 26. Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
Answer:
Let the length be l m and breadth b m. Perimeter \( 2(l + b) = 80 \Rightarrow b = 40 - l \).
Area \( = l(40 - l) = 400 \Rightarrow l^2 - 40l + 400 = 0 \). \( D = (-40)^2 - 4(1)(400) = 1600 - 1600 = 0 \).
Since \( D = 0 \), the roots are real and equal, so the situation is possible: \( (l - 20)^2 = 0 \Rightarrow l = 20 \) and \( b = 40 - 20 = 20 \).
Hence it is possible: the length and breadth are both 20 m (the park is a square).
In simple words: D = 0 means exactly one solution, and the park turns out to be a square.

Teacher's Note:
1. A rectangle with the largest area for a given perimeter is a square.
2. Equal roots still give a valid situation.