CBSE Class 10 Quadratic Equations Sure Shot Questions Set I

Question. Assertion (A): The equation \( x^2 + 3x + 1 = (x – 2)^2 \) is a quadratic equation.
Reason (R): Any equation of the form \( ax^2 + bx + c = 0 \) where \( a \neq 0 \), is a quadratic equation.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (d)

Question. Assertion (A): \( (2x – 1)^2 – 4x^2 + 5 = 0 \) is not a quadratic equation.
Reason (R): \( x = 0, 3 \) are the roots of the equation \( 2x^2 – 6x = 0 \).
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (b)

Question. If \( x = -\frac{1}{2} \) is a solution of the quadratic equation \( 3x^2 + 2kx – 3 = 0 \), find the value of \( k \).
Answer: \( k = -2\frac{1}{4} \)

Question. Find the value of \( k \) for which \( x = \sqrt{3} \) is a solution of the equation \( kx^2 + \sqrt{3}x - 4 = 0 \).
Answer: \( k = \frac{1}{3} \)

Question. If \( x = \frac{2}{3} \) and \( x = -3 \) are roots of the quadratic equations \( ax^2 + 7x + b = 0 \), find the values of \( a \) and \( b \).
Answer: \( a = 3, b = -6 \)

Question. Show that \( x = -2 \) is a solution of the equation \( 3x^2 + 13x + 14 = 0 \).
Answer: Substituting \( x = -2 \) in the given equation:
\( 3(-2)^2 + 13(-2) + 14 = 3(4) - 26 + 14 = 12 - 26 + 14 = 0 \).
Since it satisfies the equation, \( x = -2 \) is a solution.

Question. John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they have now is 124. We would like to find out how many marbles they had to start with.
Answer: Let John had \( x \) marbles. Then Jivanti had \( 45 - x \) marbles.
After losing 5 marbles each:
John has \( x - 5 \) marbles.
Jivanti has \( 45 - x - 5 = 40 - x \) marbles.
Product: \( (x - 5)(40 - x) = 124 \)
\(\Rightarrow 40x - x^2 - 200 + 5x = 124 \)
\(\Rightarrow x^2 - 45x + 324 = 0 \)

Question. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was \( Rs  750 \). We would like to find out the number of toys produced on that day.
Answer: Let the number of toys produced be \( x \).
Cost of production of each toy = \( 55 - x \).
Total cost = \( x(55 - x) = 750 \)
\(\Rightarrow 55x - x^2 = 750 \)
\(\Rightarrow x^2 - 55x + 750 = 0 \)

Question. If one root of the quadratic equation \( 3x^2 + px + 4 = 0 \) is \( \frac{2}{3} \), then find the value of \( p \) and the other root of the equation.
Answer: Substituting \( x = \frac{2}{3} \):
\( 3\left(\frac{2}{3}\right)^2 + p\left(\frac{2}{3}\right) + 4 = 0 \Rightarrow \frac{4}{3} + \frac{2p}{3} + 4 = 0 \)
\(\Rightarrow 4 + 2p + 12 = 0 \Rightarrow 2p = -16 \Rightarrow p = -8 \).
Equation: \( 3x^2 - 8x + 4 = 0 \Rightarrow 3x^2 - 6x - 2x + 4 = 0 \Rightarrow (3x - 2)(x - 2) = 0 \).
Other root is \( x = 2 \).

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of \( x \text{ km/h} \) while Ajay’s car travels \( 5 \text{ km/h} \) faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.

Question. What will be the distance covered by Ajay’s car in two hours?
(a) \( 2(x + 5) \text{ km} \)
(b) \( (x – 5) \text{ km} \)
(c) \( 2(x + 10) \text{ km} \)
(d) \( (2x + 5) \text{ km} \)
Answer: (a)

Question. Which of the following quadratic equations describes the speed of Raj’s car?
(a) \( x^2 – 5x – 500 = 0 \)
(b) \( x^2 + 4x – 400 = 0 \)
(c) \( x^2 + 5x – 500 = 0 \)
(d) \( x^2 – 4x + 400 = 0 \)
Answer: (c)

Question. The roots of the quadratic equation which describe the speed of Raj’s car are
(a) 15, -20
(b) 20, -15
(c) 20, -25
(d) 25, -25
Answer: (c)

Question. Which of the following quadratic equations has 2 as a root?
(a) \( x^2 – 4x + 5 = 0 \)
(b) \( x^2 + 3x – 12 = 0 \)
(c) \( 2x^2 – 7x + 6 = 0 \)
(d) \( 3x^2 – 6x – 2 = 0 \)
Answer: (c)

Question. The positive root of \( \sqrt{3x^2 + 6} = 9 \) is
(a) 5
(b) -5
(c) 3
(d) -3
Answer: (a)

Solution of a Quadratic Equation by Factorisation

To find the solution of a quadratic equation by factorisation method, we first express the given equation as product of two linear factors by splitting the middle term. By equating each factor to zero, we get possible solutions/roots of the given quadratic equation.
Let the given quadratic equation be \( ax^2 + bx + c = 0 \). Let the quadratic polynomial \( ax^2 + bx + c \) be expressed as the product of two linear factors say \( (px + q) \) and \( (rx + s) \) where, \( p, q, r, s \) are real numbers such that \( p \neq 0, r \neq 0 \).
Then \( ax^2 + bx + c = 0 \Rightarrow (px + q)(rx + s) = 0 \)
\(\Rightarrow\) Either \( (px + q) = 0 \) or \( (rx + s) = 0 \)
\(\Rightarrow x = -\frac{q}{p} \) or \( x = -\frac{s}{r} \)

Question. Solve the following quadratic equations by the factorisation method.
(a) \( 7x^2 = 8 – 10x \)
(b) \( x (x + 9) = 52 \)
(c) \( 3 (x^2 – 4) = 5x \)
(d) \( x (x + 1) + (x + 2) (x + 3) = 42 \)
(e) \( 3x^2 – 2\sqrt{6}x + 2 = 0 \)
Answer:
(a) \( 7x^2 + 10x - 8 = 0 \Rightarrow 7x^2 + 14x - 4x - 8 = 0 \Rightarrow (7x - 4)(x + 2) = 0 \). Hence \( x = \frac{4}{7}, -2 \).
(b) \( x^2 + 9x - 52 = 0 \Rightarrow x^2 + 13x - 4x - 52 = 0 \Rightarrow (x + 13)(x - 4) = 0 \). Hence \( x = -13, 4 \).
(c) \( 3x^2 - 5x - 12 = 0 \Rightarrow 3x^2 - 9x + 4x - 12 = 0 \Rightarrow (3x + 4)(x - 3) = 0 \). Hence \( x = -\frac{4}{3}, 3 \).
(d) \( 2x^2 + 6x - 36 = 0 \Rightarrow x^2 + 3x - 18 = 0 \Rightarrow (x + 6)(x - 3) = 0 \). Hence \( x = -6, 3 \).
(e) \( 3x^2 - 2\sqrt{6}x + 2 = 0 \Rightarrow (\sqrt{3}x - \sqrt{2})^2 = 0 \). Hence \( x = \sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}} \).

Question. Solve the equation \( \frac{4}{x} - 3 = \frac{5}{2x + 3} \); \( x \neq 0, -\frac{3}{2} \).
Answer: \( x = 1 \) or \( x = -2 \)

Question. Solve for \( x \): \( \frac{16}{x} - 1 = \frac{15}{x + 1} \); \( x \neq 0, -1 \).
Answer: \( x = \pm 4 \)

Question. Solve for \( x \): \( \frac{x-4}{x-5} + \frac{x-6}{x-7} = \frac{10}{3} \); \( x \neq 5, 7 \).
Answer: \( x = 8 \) or \( x = \frac{11}{2} \)

Question. The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Answer: 13 and 15.

Question. In the centre of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be \( 1184 \text{ m}^2 \). Find the length and breadth of the pond.
Answer: Width of grass area \( x = 8 \text{ m} \). Length of pond = 34 m, breadth of pond = 24 m.

Question. The difference of two natural numbers is 5 and the difference of their reciprocals is \( \frac{1}{10} \). Find the numbers.
Answer: 10 and 5.

Question. The sum of two numbers is 15 and the sum of their reciprocals is \( \frac{3}{10} \). Find the numbers.
Answer: 10 and 5.

Question. If Zeba was younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?
Answer: 14 years.

Question. Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream.
Answer: \( 5 \text{ km/hr} \)

Question. A train travelling at a uniform speed for 360 km would have taken 48 minutes less to travel the same distance if its speed were 5 km/hour more. Find the original speed of the train.
Answer: \( 45 \text{ km/h} \)

Question. A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.
Answer: \( 500 \text{ km/hr} \)

Question. Two pipes running together can fill a cistern in \( 3\frac{1}{13} \) hours. If one pipe takes 3 hours more than the other to fill it, find the time in which each pipe would fill the cistern.
Answer: Faster pipe takes 5 hours and slower pipe takes 8 hours respectively.

Question. Solve for \( x \): \( \frac{1}{x+1} + \frac{3}{5x+1} = \frac{5}{x+4} \), \( x \neq -1, -\frac{1}{5}, -4 \).
Answer: \( x = 1, -\frac{11}{17} \)

Exercise

Question. The roots of the equation \( \frac{4}{3}x^2 - 2x + \frac{3}{4} = 0 \) are
(a) \( \frac{2}{3}, \frac{3}{2} \)
(b) \( \frac{3}{4}, \frac{3}{4} \)
(c) \( \frac{1}{2}, -\frac{1}{2} \)
(d) None of these
Answer: (d)

Question. The required solution of \( 4x^2 – 25x = 0 \) are
(a) \( x = 0, x = \frac{12}{7} \)
(b) \( x = 0, x = \frac{25}{4} \)
(c) \( x = 1, x = \frac{5}{9} \)
(d) \( x = 1, x = \frac{12}{7} \)
Answer: (b)

Question. Choose the correct answer from the given options: For what value of k, the equation \( 9x^2 - 24x + k = 0 \) has equal roots?
(a) 12
(b) 16
(c) 18
(d) 20
Answer: (b)

Question. Choose the correct answer from the given options: The values of k for which the quadratic equation \( (k + 1)x^2 + 2(k - 1)x + (k - 2) = 0 \) has equal roots, is:
(a) \( k = 2 \)
(b) \( k = 3 \)
(c) \( k = 0 \)
(d) None of these
Answer: (b)

Question. Choose the correct answer from the given options: The value(s) of k for which the quadratic equation \( 2x^2 + kx + 2 = 0 \) has equal roots, is
(a) 4
(b) \( \pm 4 \)
(c) \( -4 \)
(d) 0
Answer: (b)

Question. Choose the correct answer from the given options: The value(s) of k, for which the roots of the equation \( 3x^2 + 2k + 27 = 0 \) are real and equal are
(a) \( k = 9 \)
(b) \( k = \pm 9 \)
(c) \( k = -9 \)
(d) \( k = 0 \)
Answer: (b)

Question. Choose the correct answer from the given options: The value of k, for which the equation \( 2x^2 - 10x + k = 0 \) has real roots is
(a) \( k \leq \frac{25}{2} \)
(b) \( k \geq \frac{25}{2} \)
(c) \( k = \frac{25}{2} \)
(d) \( k > \frac{25}{2} \)
Answer: (a)

Question. Choose the correct answer from the given options: If one root of the equation \( (k - 1)x^2 - 10x + 3 = 0 \) is the reciprocal of the other, then the value of k is
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (d)

Question. Choose the correct answer from the given options: If the quadratic equation \( x^2 - 2x + k = 0 \) has equal roots, then value of k is
(a) 1
(b) 2
(c) 3
(d) 0
Answer: (a)

Question. Choose the correct answer from the given options: If quadratic equation \( 3x^2 - 4x + k = 0 \) has equal roots, then the value of k is
(a) \( \frac{1}{3} \)
(b) \( \frac{2}{3} \)
(c) 3
(d) \( \frac{4}{3} \)
Answer: (d)

Question. Assertion (A): The equation \( 8x^2 + 3kx + 2 = 0 \) has equal roots, then the value of k is \( \pm \frac{8}{3} \).
Reason (R): The equation \( ax^2 + bx + c = 0 \) has equal roots if \( D = b^2 - 4ac = 0 \).
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (a)

Question. Assertion (A): The roots of the quadratic equation \( x^2 + 2x + 2 = 0 \) are imaginary.
Reason (R): If discriminant \( D = b^2 - 4ac < 0 \), then the roots of quadratic equation \( ax^2 + bx + c = 0 \) are imaginary.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (a)

Question. Find the value of p, so that the quadratic equation \( px(x - 3) + 9 = 0 \) has equal roots.
Answer: \( px(x - 3) + 9 = 0 \Rightarrow px^2 - 3px + 9 = 0 \).
When roots are equal, \( D = b^2 - 4ac = 0 \)
\( 9p^2 - 36p = 0 \Rightarrow 9p(p - 4) = 0 \Rightarrow p = 0, p = 4 \).
But \( p \neq 0 \) (In quadratic equation, \( a \neq 0 \)).
\( \therefore p = 4 \)

Question. For what values of k, the roots of the equation \( x^2 + 4x + k = 0 \) are real?
Answer: For real roots, \( D \geq 0 \Rightarrow b^2 - 4ac \geq 0 \)
\( (4)^2 - 4 \times 1 \times k \geq 0 \Rightarrow 16 - 4k \geq 0 \Rightarrow 16 \geq 4k \Rightarrow k \leq 4 \)

Question. Find the value of k for which the roots of the equation \( 3x^2 - 10x + k = 0 \) are reciprocal of each other.
Answer: Let the roots be \( \alpha \) and \( \frac{1}{\alpha} \).
Product of roots \( = \alpha \cdot \frac{1}{\alpha} = \frac{c}{a} \)
\( 1 = \frac{k}{3} \Rightarrow k = 3 \)

Question. Find the discriminant of the quadratic equation \( 2x^2 - 4x + 3 = 0 \), hence find the nature of its roots.
Answer: \( D = b^2 - 4ac = (-4)^2 - 4(2)(3) = 16 - 24 = -8 \).
Since \( D < 0 \), there are no real roots.

Question. If \( x = 3 \) is one root of the quadratic equation \( x^2 - 2kx - 6 = 0 \), then find the value of k.
Answer: Since \( x = 3 \) is a root, \( (3)^2 - 2k(3) - 6 = 0 \)
\( 9 - 6k - 6 = 0 \Rightarrow 3 - 6k = 0 \Rightarrow k = \frac{3}{6} = \frac{1}{2} \)

Question. For what values of k, the equation \( 9x^2 + 6kx + 4 = 0 \) has equal roots?
Answer: For equal roots, \( D = 0 \)
\( (6k)^2 - 4 \times 9 \times 4 = 0 \Rightarrow 36k^2 = 144 \Rightarrow k^2 = 4 \Rightarrow k = \pm 2 \)

Question. For what value(s) of 'a' quadratic equation \( 3ax^2 - 6x + 1 = 0 \) has no real roots?
Answer: For no real roots, \( D < 0 \)
\( (-6)^2 - 4(3a)(1) < 0 \Rightarrow 36 - 12a < 0 \Rightarrow 12a > 36 \Rightarrow a > 3 \)

Short Answer Type Questions-

Question. State whether the quadratic equation \( 4x^2 - 5x + \frac{25}{16} = 0 \) has two distinct real roots or not. Justify your answer.
Answer: No. For \( 4x^2 - 5x + \frac{25}{16} = 0 \), \( D = b^2 - 4ac = (-5)^2 - 4 \times 4 \times \frac{25}{16} = 25 - 25 = 0 \). Since \( D = 0 \), it has equal real roots, not distinct.

Question. Find the value of k so that the quadratic equation \( kx(3x - 10) + 25 = 0 \), has two equal roots.
Answer: \( 3kx^2 - 10kx + 25 = 0 \). For equal roots, \( D = 0 \)
\( (-10k)^2 - 4 \times 3k \times 25 = 0 \Rightarrow 100k^2 - 300k = 0 \Rightarrow 100k(k - 3) = 0 \)
Since \( k \neq 0 \), \( k = 3 \)

Question. For what value of k does the quadratic equation \( (k - 5)x^2 + 2(k - 5)x + 2 = 0 \) have equal roots?
Answer: For equal roots, \( D = 0 \)
\( [2(k - 5)]^2 - 4(k - 5)(2) = 0 \Rightarrow 4(k - 5)^2 - 8(k - 5) = 0 \)
\( 4(k - 5)[(k - 5) - 2] = 0 \Rightarrow 4(k - 5)(k - 7) = 0 \)
As \( k \neq 5 \) (otherwise it's not a quadratic equation), \( k = 7 \)

Question. Find the value(s) of k so that the quadratic equation \( 2x^2 + kx + 3 = 0 \) has equal roots.
Answer: For equal roots, \( D = 0 \Rightarrow k^2 - 4(2)(3) = 0 \Rightarrow k^2 = 24 \Rightarrow k = \pm \sqrt{24} = \pm 2\sqrt{6} \)

Question. Find the value(s) of k so that the quadratic equation \( x^2 - 4kx + k = 0 \) has equal roots.
Answer: For equal roots, \( D = 0 \Rightarrow (-4k)^2 - 4(1)(k) = 0 \Rightarrow 16k^2 - 4k = 0 \Rightarrow 4k(4k - 1) = 0 \)
\( k = 0 \) or \( k = \frac{1}{4} \)

Question. Find the value(s) of k so that the quadratic equation \( 3x^2 - 2kx + 12 = 0 \) has equal roots.
Answer: For equal roots, \( D = 0 \Rightarrow (-2k)^2 - 4(3)(12) = 0 \Rightarrow 4k^2 - 144 = 0 \Rightarrow 4k^2 = 144 \Rightarrow k^2 = 36 \Rightarrow k = \pm 6 \)

Question. Find the values of k for which the quadratic equation \( 9x^2 - 3kx + k = 0 \) has equal roots.
Answer: For equal roots, \( D = 0 \Rightarrow (-3k)^2 - 4(9)(k) = 0 \Rightarrow 9k^2 - 36k = 0 \Rightarrow 9k(k - 4) = 0 \)
Since \( k \neq 0 \), \( k = 4 \)