CBSE Class 10 Mathematics Quadratic Equations Assignment Set 27

Exercise 1: Objective Questions

Question 1. Which of the following equations has -1 as a root?
(A) \( x^2 + 3x - 10 = 0 \)   (B) \( x^2 - x - 12 = 0 \)   (C) \( 3x^2 - 2x - 5 = 0 \)   (D) \( 9x^2 + 24x + 16 = 0 \)
Answer:
Ans. (C) \( 3x^2 - 2x - 5 = 0 \)
Put x = -1: (A) 1 - 3 - 10 = -12; (B) 1 + 1 - 12 = -10; (C) 3 + 2 - 5 = 0; (D) 9 - 24 + 16 = 1. Only (C) gives 0.

Teacher's Note:
1. Substitute the value in each option.
2. A root makes the equation zero.

Question 2. The value of k for which 3 is a root of \( kx^2 - 7x + 3 = 0 \) is
(A) 2   (B) -2   (C) 3   (D) -3
Answer:
Ans. (A) 2
\( 9k - 21 + 3 = 0 \Rightarrow 9k = 18 \Rightarrow k = 2 \).

Teacher's Note:
1. Substitute x = 3.
2. Solve the linear equation for k.

Question 3. The roots of the equation \( x^2 - 9x + 20 = 0 \) are
(A) 4 and 5   (B) -4 and 5   (C) 4 and -5   (D) -4 and -5
Answer:
Ans. (A) 4 and 5
\( x^2 - 9x + 20 = (x - 4)(x - 5) \), so x = 4 or 5.

Teacher's Note:
1. Find two numbers with sum 9 and product 20.
2. Both are positive.

Question 4. If one root of \( 5x^2 + 13x + k = 0 \) is the reciprocal of the other, then k =
(A) 0   (B) 5   (C) \( \tfrac{1}{6} \)   (D) 6
Answer:
Ans. (B) 5
Product of roots \( = \dfrac{k}{5} = 1 \Rightarrow k = 5 \).

Teacher's Note:
1. Reciprocal roots have product 1.
2. Product = c/a.

Question 5. Which of the following equations has the sum of its roots equal to 2?
(A) \( 5x^2 - \tfrac{2}{5}x - 3 = 0 \)   (B) \( x^2 - 2x - 5 = 0 \)   (C) \( 2x^2 - 2x + 1 = 0 \)   (D) \( 3x^2 - 2x + 7 = 0 \)
Answer:
Ans. (B) \( x^2 - 2x - 5 = 0 \)
Sum of roots \( = -\dfrac{b}{a} \): (A) \( \tfrac{2}{25} \); (B) 2; (C) 1; (D) \( \tfrac{2}{3} \). Only (B) gives 2.

Teacher's Note:
1. Use sum = -b/a.
2. Check each option.

Question 6. The sum of the reciprocals of the roots of the equation \( \dfrac{2009}{2010}x + 1 + \dfrac{1}{x} = 0 \) is
(A) \( -\tfrac{2010}{2009} \)   (B) -1   (C) \( \tfrac{2009}{2010} \)   (D) 1
Answer:
Ans. (B) -1
Multiplying by x (x is not 0): \( \dfrac{2009}{2010}x^2 + x + 1 = 0 \). Sum of reciprocals \( = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{-b/a}{c/a} = -\dfrac{b}{c} = -1 \).

Teacher's Note:
1. Sum of reciprocals = -b/c.
2. Clear the fraction first.

Question 7. A quadratic equation \( ax^2 - 2ax + b = 0 \) has two real solutions. Their average is
(A) 1   (B) 2   (C) \( \tfrac{b}{a} \)   (D) \( \tfrac{2b}{a} \)
Answer:
Ans. (A) 1
Sum of roots \( = \dfrac{2a}{a} = 2 \), so the average is 1.

Teacher's Note:
1. Average = sum / 2.
2. Sum = -b/a.

Question 8. The roots of the equation \( x^2 - x - 3 = 0 \) are
(A) imaginary   (B) rational   (C) irrational   (D) none of these
Answer:
Ans. (C) irrational
\( D = 1 + 12 = 13 \), positive but not a perfect square. The roots are real and irrational.

Teacher's Note:
1. Compute D.
2. Perfect square gives rational roots.

Question 9. If one of the roots of a quadratic equation having rational coefficients is \( 2 + \sqrt{3} \), then the equation is
(A) \( x^2 - (2 + \sqrt{3})x + 1 = 0 \)   (B) \( x^2 + (2 + \sqrt{3})x + 1 = 0 \)   (C) \( x^2 - 4x + 1 = 0 \)   (D) \( x^2 + 4x - 1 = 0 \)
Answer:
Ans. (C) \( x^2 - 4x + 1 = 0 \)
With rational coefficients the other root is \( 2 - \sqrt{3} \). Sum = 4, product = 4 - 3 = 1, so \( x^2 - 4x + 1 = 0 \).

Teacher's Note:
1. Irrational roots come in conjugate pairs.
2. Form x² - (sum)x + product.

Question 10. Which of the following equations has two equal real roots?
(A) \( 3x^2 + 14x - 5 = 0 \)   (B) \( 4x^2 + 2x - 1 = 0 \)   (C) \( 9x^2 - 6x + 1 = 0 \)   (D) \( x^2 - 5x + 4 = 0 \)
Answer:
Ans. (C) \( 9x^2 - 6x + 1 = 0 \)
For \( 9x^2 - 6x + 1 \): \( D = 36 - 36 = 0 \). The others have D = 256, 20 and 9, all positive.

Teacher's Note:
1. Equal roots need D = 0.
2. \( 9x^2 - 6x + 1 = (3x - 1)^2 \).

Question 11. If the equation \( 16x^2 + 6kx + 4 = 0 \) has equal roots, then the value of k is
(A) \( \pm 8 \)   (B) \( \pm\tfrac{8}{3} \)   (C) \( \pm\tfrac{3}{8} \)   (D) 0
Answer:
Ans. (B) \( \pm\tfrac{8}{3} \)
\( D = 36k^2 - 256 = 0 \Rightarrow k^2 = \dfrac{64}{9} \Rightarrow k = \pm\dfrac{8}{3} \).

Teacher's Note:
1. Set D = 0.
2. Take both signs.

Question 12. The quadratic equation \( x^2 - 5x + 3 = 0 \) has
(A) no real roots   (B) two distinct real roots   (C) two equal real roots   (D) more than two real roots
Answer:
Ans. (B) two distinct real roots
\( D = 25 - 12 = 13 > 0 \): two distinct real roots.

Teacher's Note:
1. D > 0 gives two distinct real roots.
2. A quadratic never has more than two roots.

Question 13. The difference between two numbers is 5 and the difference between their squares is 65. The larger number is
(A) 9   (B) 10   (C) 11   (D) 12
Answer:
Ans. (A) 9
\( (x - y)(x + y) = 65 \) with \( x - y = 5 \) gives \( x + y = 13 \). So x = 9 and y = 4.

Teacher's Note:
1. Use a² - b² = (a - b)(a + b).
2. Solve the two linear equations.

Question 14. The sum of the ages of a father and his son is 45 years. Five years ago the product of their ages was 4 times the age of the father at that time. The present age of the father is
(A) 30 years   (B) 31 years   (C) 36 years   (D) 41 years
Answer:
Ans. (C) 36 years
Let the father's age be f and the son's age be 45 - f. Five years ago: \( (f - 5)(40 - f) = 4(f - 5) \). Since \( f \ne 5 \), \( 40 - f = 4 \), so f = 36 (son 9).

Teacher's Note:
1. Cancel the common factor (f - 5).
2. Check: 31 x 4 = 4 x 31.

Question 15. The sum of the squares of two consecutive positive integers is 545. The sum of these integers is
(A) 33   (B) 34   (C) 35   (D) 36
Answer:
Ans. (A) 33
\( n^2 + (n + 1)^2 = 545 \Rightarrow n^2 + n - 272 = 0 \Rightarrow (n - 16)(n + 17) = 0 \). n = 16, so the integers are 16 and 17 and the sum is 33.

Teacher's Note:
1. Let the integers be n and n + 1.
2. Reject the negative root.

Question 16. The sides (in cm) of a right triangle containing the right angle are 5x and 3x - 1. If the area of the triangle is 60 cm², its perimeter is
(A) 30 cm   (B) 35 cm   (C) 40 cm   (D) 45 cm
Answer:
Ans. (C) 40 cm
\( \tfrac{1}{2}(5x)(3x - 1) = 60 \Rightarrow 3x^2 - x - 24 = 0 \Rightarrow (x - 3)(3x + 8) = 0 \), so x = 3. The legs are 15 and 8, the hypotenuse is 17 and the perimeter is 40 cm.

Teacher's Note:
1. Area = half the product of the legs.
2. 8-15-17 is a Pythagorean triple.

Question 17. A two-digit number is four times the sum of its digits and three times the product of its digits. The number is
(A) 24   (B) 42   (C) 36   (D) 63
Answer:
Ans. (A) 24
Let the number be 10a + b. \( 10a + b = 4(a + b) \Rightarrow b = 2a \); \( 10a + b = 3ab \Rightarrow 12a = 6a^2 \Rightarrow a = 2 \), so b = 4. The number is 24.

Teacher's Note:
1. Use both conditions to relate a and b.
2. Check: 4 x 6 = 24 and 3 x 8 = 24.

Exercise 2: Objective Questions

Question 18. If the roots of the equation \( px^2 + qx + r = 0 \) are in the ratio l : m, then
(A) \( (l + m)^2 pq = lmr^2 \)   (B) \( (l + m)^2 pr = lmq \)   (C) \( (l + m)^2 pr = lmq^2 \)   (D) none of these
Answer:
Ans. (C) \( (l + m)^2 pr = lmq^2 \)
Let the roots be l t and m t. Sum \( (l + m)t = -\dfrac{q}{p} \), product \( lmt^2 = \dfrac{r}{p} \). Eliminating t: \( \dfrac{lm}{(l + m)^2} \cdot \dfrac{q^2}{p^2} = \dfrac{r}{p} \), so \( (l + m)^2 pr = lmq^2 \).

Teacher's Note:
1. Write the roots as lt and mt.
2. Eliminate t using sum and product.

Question 19. If \( 2 - \sqrt{3} \) is a root of \( x^2 + 2(\sqrt{3} - 1)x + 3 - 2\sqrt{3} = 0 \), then the second root is
(A) \( \sqrt{3} - 2 \)   (B) \( \sqrt{3} \)   (C) \( 2 + \sqrt{3} \)   (D) \( -\sqrt{3} \)
Answer:
Ans. (D) \( -\sqrt{3} \)
Sum of roots \( = -2(\sqrt{3} - 1) = 2 - 2\sqrt{3} \). Second root \( = 2 - 2\sqrt{3} - (2 - \sqrt{3}) = -\sqrt{3} \). (Check: product \( = (2 - \sqrt{3})(-\sqrt{3}) = 3 - 2\sqrt{3} \).)

Teacher's Note:
1. Use the sum of roots.
2. Verify with the product.

Question 20. \( -2 + \sqrt{3} \) is a solution of which of the following equations?
(A) \( 3x^2 + 12x + 3 = 0 \)   (B) \( x^2 + 2x + 3 = 0 \)   (C) \( x^2 + 4x + 2 = 0 \)   (D) \( 3x^2 + 2x - 1 = 0 \)
Answer:
Ans. (A) \( 3x^2 + 12x + 3 = 0 \)
Let \( x = -2 + \sqrt{3} \); then \( x + 2 = \sqrt{3} \), so \( (x + 2)^2 = 3 \), i.e. \( x^2 + 4x + 1 = 0 \). Multiplying by 3: \( 3x^2 + 12x + 3 = 0 \), option (A).

Teacher's Note:
1. Isolate the surd and square.
2. Match with the options.

Question 21. The value(s) of k for which the quadratic equation \( kx^2 - kx + 1 = 0 \) has equal roots is
(A) k = 0 only   (B) k = 4 only   (C) k = 0, 4   (D) k = -4
Answer:
Ans. (B) k = 4 only
\( D = k^2 - 4k = 0 \Rightarrow k = 0 \) or 4. For k = 0 the equation becomes 1 = 0, which is not quadratic. So k = 4 only.

Teacher's Note:
1. The coefficient of x² must be non-zero.
2. Reject k = 0.

Question 22. An aeroplane left 30 minutes later than its scheduled time and, in order to reach its destination 1500 km away in time, it has to increase its speed by 250 km/h from its usual speed. Its usual speed is
(A) 700 km/h   (B) 750 km/h   (C) 770 km/h   (D) 800 km/h
Answer:
Ans. (B) 750 km/h
\( \dfrac{1500}{x} - \dfrac{1500}{x + 250} = \dfrac{1}{2} \Rightarrow x^2 + 250x - 750000 = 0 \Rightarrow (x + 1000)(x - 750) = 0 \). x = 750 km/h.

Teacher's Note:
1. Convert 30 minutes to 1/2 hour.
2. Reject the negative root.

Question 23. Two water taps together can fill a tank in \( 9\dfrac{3}{8} \) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. The times taken by the taps separately are
(A) 25 h, 15 h   (B) 15 h, 5 h   (C) 35 h, 25 h   (D) 10 h, 20 h
Answer:
Ans. (A) 25 h, 15 h
Let the larger tap take x hours and the smaller x + 10: \( \dfrac{1}{x} + \dfrac{1}{x + 10} = \dfrac{8}{75} \Rightarrow 4x^2 - 35x - 375 = 0 \Rightarrow (x - 15)(4x + 25) = 0 \). x = 15 h, so the smaller tap takes 25 h.

Teacher's Note:
1. Total time 75/8 hours means rate 8/75.
2. Reject the negative root.

Question 24. A girl is twice as old as her sister. Four years hence, the product of their ages (in years) will be 160. The present age of her sister is
(A) 12 years   (B) 6 years   (C) 8 years   (D) 9 years
Answer:
Ans. (B) 6 years
Sister's age s: \( (2s + 4)(s + 4) = 160 \Rightarrow s^2 + 6s - 72 = 0 \Rightarrow (s + 12)(s - 6) = 0 \). s = 6.

Teacher's Note:
1. Add 4 to both ages.
2. Reject the negative root.

Question 25. One of two students, while solving a quadratic equation in x, copied the constant term incorrectly and got the roots 3 and 2. The other copied the constant term and the coefficient of \( x^2 \) correctly as -6 and 1 respectively. The correct roots are
(A) 3, -2   (B) -3, 2   (C) -6, -1   (D) 6, -1
Answer:
Ans. (D) 6, -1
The first student's sum of roots 5 is correct (only the constant was wrong), so b = -5. The correct equation is \( x^2 - 5x - 6 = 0 \Rightarrow (x - 6)(x + 1) = 0 \): roots 6 and -1.

Teacher's Note:
1. The sum of the roots depends on the x² and x coefficients only.
2. Use the second student's constant term.

Question 26. Out of a number of Saras birds, one fourth of the number are moving about in lots, \( \dfrac{1}{9} \)th along with \( \dfrac{1}{4} \)th as well as 7 times the square root of the number move on a hill, and 56 birds remain in Vakula trees. The total number of birds is
(A) 576   (B) 500   (C) 676   (D) 400
Answer:
Ans. (A) 576
\( n = \dfrac{n}{4} + \dfrac{n}{9} + \dfrac{n}{4} + 7\sqrt{n} + 56 \Rightarrow \dfrac{7n}{18} - 7\sqrt{n} - 56 = 0 \Rightarrow n - 18\sqrt{n} - 144 = 0 \). With \( t = \sqrt{n} \): \( t^2 - 18t - 144 = 0 \Rightarrow t = 24 \) (t > 0). So n = 576.

Teacher's Note:
1. Let the square root of n be t.
2. Check: 144 + 64 + 144 + 168 + 56 = 576.

Question 27. If n is an integer such that \( n \times n = n + n \), then the number of such n is
(A) 0   (B) 1   (C) 2   (D) 3
Answer:
Ans. (C) 2
\( n^2 = 2n \Rightarrow n(n - 2) = 0 \Rightarrow n = 0 \) or n = 2. Both are integers, so there are 2 such values of n.

Teacher's Note:
1. Solve n² = 2n without dividing by n.
2. Dividing by n would lose the root n = 0.

Question 28. A positive number whose reciprocal equals one less than the number is
(A) \( \dfrac{1 + \sqrt{2}}{2} \)   (B) \( \dfrac{\sqrt{2} - 1}{4} \)   (C) \( \dfrac{1 + \sqrt{5}}{2} \)   (D) \( \dfrac{\sqrt{2} + \sqrt{5}}{2} \)
Answer:
Ans. (C) \( \dfrac{1 + \sqrt{5}}{2} \)
\( \dfrac{1}{x} = x - 1 \Rightarrow x^2 - x - 1 = 0 \Rightarrow x = \dfrac{1 \pm \sqrt{5}}{2} \). The positive value is \( \dfrac{1 + \sqrt{5}}{2} \).

Teacher's Note:
1. Form the quadratic and use the formula.
2. Reject the negative root.

Question 29. If \( \alpha \) is a root, repeated twice, of the quadratic equation \( (a - d)x^2 + ax + (a + d) = 0 \), then \( \dfrac{d^2}{a^2} \) has the value equal to
(A) \( \sin^2 90^\circ \)   (B) \( \cos^2 60^\circ \)   (C) \( \sin^2 45^\circ \)   (D) \( \cos^2 30^\circ \)
Answer:
Ans. (D) \( \cos^2 30^\circ \)
Equal roots: \( a^2 - 4(a - d)(a + d) = 0 \Rightarrow 4d^2 = 3a^2 \Rightarrow \dfrac{d^2}{a^2} = \dfrac{3}{4} = \cos^2 30^\circ \).

Teacher's Note:
1. A repeated root means D = 0.
2. cos 30° = √3/2.

Question 30. The quadratic equation whose roots are half of the reciprocals of the roots of \( ax^2 + bx + c = 0 \) is
(A) \( 4ax^2 + 2bx + c = 0 \)   (B) \( 4cx^2 + 2bx + a = 0 \)   (C) \( 2cx^2 + bx + a = 0 \)   (D) \( 2ax^2 + bx + c = 0 \)
Answer:
Ans. (B) \( 4cx^2 + 2bx + a = 0 \)
If \( y = \dfrac{1}{2\alpha} \) then \( \alpha = \dfrac{1}{2y} \). Substituting in \( a\alpha^2 + b\alpha + c = 0 \): \( \dfrac{a}{4y^2} + \dfrac{b}{2y} + c = 0 \Rightarrow 4cy^2 + 2by + a = 0 \).

Teacher's Note:
1. Express the old root in terms of the new root.
2. Substitute and clear fractions.

Question 31. If \( \sin a \) and \( \cos a \) are the roots of the equation \( 4x^2 - kx - 1 = 0 \) (k > 0), then the value of k is
(A) \( 2\sqrt{2} \)   (B) 4   (C) 2   (D) \( 4\sqrt{2} \)
Answer:
Ans. (A) \( 2\sqrt{2} \)
\( \sin a + \cos a = \dfrac{k}{4} \) and \( \sin a \cos a = -\dfrac{1}{4} \). Squaring the sum: \( 1 + 2\left(-\dfrac{1}{4}\right) = \dfrac{k^2}{16} \Rightarrow \dfrac{1}{2} = \dfrac{k^2}{16} \Rightarrow k^2 = 8 \), so \( k = 2\sqrt{2} \).

Teacher's Note:
1. Use sin²a + cos²a = 1.
2. Take k > 0.

Question 32. The roots of the equation \( x^2 + px + q = 0 \) are 1 and 2. The roots of the equation \( qx^2 - px + 1 = 0 \) must be
(A) \( -1, -\tfrac{1}{2} \)   (B) \( \tfrac{1}{2}, 1 \)   (C) \( -\tfrac{1}{2}, 1 \)   (D) \( -1, \tfrac{1}{2} \)
Answer:
Ans. (A) \( -1, -\tfrac{1}{2} \)
p = -(1 + 2) = -3 and q = 2. The second equation is \( 2x^2 + 3x + 1 = 0 \Rightarrow (2x + 1)(x + 1) = 0 \): roots -1 and \( -\tfrac{1}{2} \).

Teacher's Note:
1. Find p and q first.
2. Then solve the new equation.