CBSE Class 10 Quadratic Equations Sure Shot Questions Set 10

I. Very Short Answer Type Questions

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: (b)

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. Assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(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.

Assertion (A): When the quadratic equation \( 6x^2 - x - 2 = 0 \) is factorised, we get its roots as \( \frac{2}{3} \) and \( -\frac{1}{2} \).
Reason (R): \( 6x^2 - x - 2 = 0 \Rightarrow 2x(3x - 2) + (3x - 2) = 0 \Rightarrow (3x - 2)(2x + 1) = 0 \Rightarrow x = \frac{2}{3}, -\frac{1}{2} \).
Answer: (a)

Question. Assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(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.

Assertion (A): If \( x^2 - (\sqrt{3} + 1)x + \sqrt{3} = 0 \), then \( x^2 - \sqrt{3}x - x + \sqrt{3} = 0 \Rightarrow x(x - \sqrt{3}) - 1(x - \sqrt{3}) = 0 \Rightarrow (x - \sqrt{3})(x - 1) = 0 \Rightarrow x = \sqrt{3}, 1 \).
Reason (R): If we can factorise \( ax^2 + bx + c, a \neq 0 \) into a product of two linear factors, then the roots of the quadratic equation \( ax^2 + bx + c = 0 \) can be found by equating each factor to zero.
Answer: (a)

Question. Find the roots of the following quadratic equations by factorisation:
(1) \( \sqrt{3}x^2 + 10x + 7\sqrt{3} = 0 \)
(2) \( (x - 3)(2x + 3) = 0 \)
(3) \( 3x^2 - 2ax - a^2 = 0 \)
(4) \( 3a^2x^2 + 8abx + 4b^2 = 0 \)
(5) Find the roots of the equation \( x^2 + 7x + 10 = 0 \).
Answer: (1) \( -\sqrt{3}, -\frac{7}{\sqrt{3}} \); (2) \( 3, -\frac{3}{2} \); (3) \( a, -\frac{a}{3} \); (4) \( -\frac{2b}{a}, -\frac{2b}{3a} \); (5) \( -2, -5 \)

II. Short Answer Type Questions-I

Question. Solve for \( x \): \( 4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0 \).
Answer: \( \frac{\sqrt{3}}{4}, -\frac{2}{\sqrt{3}} \)

Question. Solve for \( x \): \( x^2 - (\sqrt{2} + 1)x + \sqrt{2} = 0 \).
Answer: \( \sqrt{2}, 1 \)

Question. Solve for \( x \): \( \sqrt{2x+9} + x = 13 \).
Answer: \( x = 8 \) (\( x = 20 \) is extraneous)

Question. Solve for \( x \): \( \sqrt{6x+7} - (2x-7) = 0 \).
Answer: \( x = 7 \) (\( x = 1.5 \) is extraneous)

Question. Solve for \( x \): \( \sqrt{3}x^2 - 2\sqrt{2}x - 2\sqrt{3} = 0 \).
Answer: \( \sqrt{6}, -\sqrt{\frac{2}{3}} \)

Question. Solve for \( x \): \( \frac{1}{x-3} - \frac{1}{x+5} = \frac{1}{6}, x \neq 3, -5 \).
Answer: \( x = 7, -9 \)

Question. Solve for \( x \): \( \sqrt{3}x^2 + 14x - 5\sqrt{3} = 0 \).
Answer: \( \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}, -5\sqrt{3} \)

III. Short Answer Type Questions-II

Question. Solve for \( x \): \( \frac{x+1}{x-1} + \frac{x-2}{x+2} = 4 - \frac{2x+3}{x-2}; x \neq 1, -2, 2 \).
Answer: \( x = 5, - \frac{6}{5} \)

Question. The difference of two natural numbers is 3 and the difference of their reciprocals is \( \frac{3}{28} \). Find the numbers.
Answer: 4 and 7

Question. The difference of two natural numbers is 5 and the difference of their reciprocals is \( \frac{5}{14} \). Find the numbers.
Answer: 2 and 7

IV. Long Answer Type Questions

Question. Solve the equation for \( x \): \( \frac{3x-4}{7} + \frac{7}{3x-4} = \frac{5}{2}, x \neq \frac{4}{3} \).
Answer: \( x = 6, \frac{5}{2} \)

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

Question. Some students planned a picnic. The total budget for food was ₹2,000. But 5 students failed to attend the picnic and thus the cost of food for each member increased by ₹20. How many students attended the picnic and how much did each student pay for the food?
Answer: 20 students attended, each paid ₹100

Question. A two-digit number is such that the product of its digits is 14. When 45 is added to the number, the digits interchange their places. Find the number.
Answer: 27

Question. Two water taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Answer: 9 hours and 18 hours

Question. Two pipes running together can fill a tank in \( 11\frac{1}{9} \) minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately.
Answer: 20 minutes and 25 minutes

Question. A pole has to be erected at a point on the boundary of a circular park of diameter 17 m in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Find the distances from the two gates where the pole is to be erected.
Answer: 8 m and 15 m

Question. A motorboat whose speed in still water is 18 km/h, takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Answer: 6 km/h

Question. At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.
Answer: Asha: 27 years, Nisha: 5 years

Question. A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed?
Answer: 42 km/hr

Question. Solve the following equation: \( \frac{1}{x} - \frac{1}{x-2} = 3, x \neq 0, 2 \).
Answer: \( \frac{3 \pm \sqrt{3}}{3} \)

Question. Find two consecutive positive integers sum of whose squares is 365.
Answer: 13 and 14

Question. A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m. Find the length and breadth of the park.
Answer: Length: 7 m, Breadth: 4 m

Question. In a flight of 600 km, an aircraft was slowed down due to bad weather. The average speed of the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. Find the duration of flight.
Answer: 1 hour (Original speed was 600 km/hr)

Case Study Based Questions

The speed of a motor boat is 20 km/hr. For covering the distance of 15 km the boat took 1 hour more for upstream than downstream.

Question. Let speed of the stream be \( x \) km/hr, then speed of the motorboat in upstream will be
(a) 20 km/hr
(b) \( (20 + x) \) km/hr
(c) \( (20 - x) \) km/hr
(d) 2 km/hr
Answer: (c)

Question. What is the relation between speed, distance and time?
(a) \( \text{speed} = \frac{\text{distance}}{\text{time}} \)
(b) \( \text{distance} = \frac{\text{speed}}{\text{time}} \)
(c) \( \text{time} = \text{speed} \times \text{distance} \)
(d) \( \text{speed} = \text{distance} \times \text{time} \)
Answer: (a)

Question. Which is the correct quadratic equation for the speed of the stream?
(a) \( x^2 + 30x - 200 = 0 \)
(b) \( x^2 + 20x - 400 = 0 \)
(c) \( x^2 + 30x - 400 = 0 \)
(d) \( x^2 - 20x - 400 = 0 \)
Answer: (c)

Question. What is the speed of stream?
(a) 20 km/hour
(b) 10 km/hour
(c) 15 km/hour
(d) 25 km/hour
Answer: (b)

Question. How much time boat took in downstream?
(a) 90 minutes
(b) 15 minutes
(c) 30 minutes
(d) 45 minutes
Answer: (c)

Short Answer Type Questions

Question. Find the nature of the roots of the following quadratic equations. If the real roots exist, then also find them.
(a) \( 4x^2 + 12x + 9 = 0 \)
(b) \( 3x^2 + 5x - 7 = 0 \)
(c) \( 7y^2 - 4y + 5 = 0 \)
Answer: (a) \( D = (12)^2 - 4(4)(9) = 144 - 144 = 0 \). Real and equal roots: \( x = -\frac{12}{8} = -\frac{3}{2} \).
(b) \( D = (5)^2 - 4(3)(-7) = 25 + 84 = 109 > 0 \). Real and distinct roots: \( x = \frac{-5 \pm \sqrt{109}}{6} \).
(c) \( D = (-4)^2 - 4(7)(5) = 16 - 140 = -124 < 0 \). No real roots.

Question. If 2 is a root of the quadratic equation \( 3x^2 + px - 8 = 0 \) and the quadratic equation \( 4x^2 - 2px + k = 0 \) has equal roots, find the value of k.
Answer: Since 2 is a root of \( 3x^2 + px - 8 = 0 \), \( 3(2)^2 + p(2) - 8 = 0 \Rightarrow 12 + 2p - 8 = 0 \Rightarrow 2p = -4 \Rightarrow p = -2 \).
Now, the equation \( 4x^2 - 2(-2)x + k = 0 \Rightarrow 4x^2 + 4x + k = 0 \) has equal roots.
\( D = (4)^2 - 4(4)(k) = 0 \Rightarrow 16 - 16k = 0 \Rightarrow k = 1 \).

Question. Find the value of p for which the quadratic equation \( (p + 1)x^2 - 6(p + 1)x + 3(p + 9) = 0, p \neq -1 \) has equal roots. Hence, find the roots of the equation.
Answer: \( p = 3 \). Roots are \( 3, 3 \).

Question. Find that non-zero value of k, for which the quadratic equation \( kx^2 + 1 - 2(k - 1)x + x^2 = 0 \) has equal roots. Hence, find the roots of the equation.
Answer: \( (k + 1)x^2 - 2(k - 1)x + 1 = 0 \). For equal roots, \( D = 0 \)
\( [-2(k - 1)]^2 - 4(k + 1)(1) = 0 \Rightarrow 4(k^2 - 2k + 1) - 4k - 4 = 0 \Rightarrow 4k^2 - 8k + 4 - 4k - 4 = 0 \)
\( 4k^2 - 12k = 0 \Rightarrow 4k(k - 3) = 0 \). Since \( k \neq 0 \), \( k = 3 \). Roots are \( x = \frac{1}{2} \).

Question. The roots \( \alpha \) and \( \beta \) of the quadratic equation \( x^2 - 5x + 3(k - 1) = 0 \) are such that \( \alpha - \beta = 1 \). Find the value k.
Answer: \( \alpha + \beta = 5 \). Given \( \alpha - \beta = 1 \). Adding gives \( 2\alpha = 6 \Rightarrow \alpha = 3, \beta = 2 \).
Product of roots \( \alpha\beta = 3 \times 2 = 6 \).
\( 3(k - 1) = 6 \Rightarrow k - 1 = 2 \Rightarrow k = 3 \).

Question. Find the values of k for which the quadratic equation \( (3k + 1)x^2 + 2(k + 1)x + 1 = 0 \) has equal roots. Also find these roots.
Answer: For equal roots, \( D = 0 \Rightarrow [2(k+1)]^2 - 4(3k+1)(1) = 0 \Rightarrow 4(k^2 + 2k + 1) - 12k - 4 = 0 \)
\( 4k^2 + 8k + 4 - 12k - 4 = 0 \Rightarrow 4k^2 - 4k = 0 \Rightarrow 4k(k - 1) = 0 \)
\( k = 0 \) or \( k = 1 \). Roots are \( 1 \) or \( 0 \).

Long Answer Type Questions

Question. Find whether the equation \( \frac{1}{2x-3} + \frac{1}{x-5} = 1, x \neq \frac{3}{2}, 5 \) has real roots. If real roots exist, find them.
Answer: Yes. After simplifying, the equation becomes \( x^2 - 10x + 23 = 0 \). \( D = 100 - 92 = 8 > 0 \). Roots are \( x = \frac{10 \pm \sqrt{8}}{2} = \frac{10 \pm 2\sqrt{2}}{2} = 5 \pm \sqrt{2} \).

Question. Check whether the equation \( 5x^2 - 6x - 2 = 0 \) has real roots and if it has, find them by the method of completing the square. Also verify that roots obtained satisfy the given equation.
Answer: \( D = (-6)^2 - 4(5)(-2) = 36 + 40 = 76 > 0 \). Roots are real.
\( 5x^2 - 6x - 2 = 0 \Rightarrow x^2 - \frac{6}{5}x = \frac{2}{5} \Rightarrow x^2 - \frac{6}{5}x + \left(\frac{3}{5}\right)^2 = \frac{2}{5} + \frac{9}{25} \)
\( \left(x - \frac{3}{5}\right)^2 = \frac{10 + 9}{25} = \frac{19}{25} \Rightarrow x - \frac{3}{5} = \pm \frac{\sqrt{19}}{5} \)
\( x = \frac{3 \pm \sqrt{19}}{5} \)