Previous Year Questions: Quadratic Equations

# Previous Year Questions: Quadratic Equations - Mathematics (Maths) Class 10

## Short Answer Type Questions

Q.1. What will be the nature of roots of quadratic equation 2x2 - 4x + 3 = 0?

[CBSE 2019 ]
Ans:
D = b2 - 4ac = 42 - 4 x 2 x 3 = 16 - 24 = -8 < 0
Since D < 0
Hence, roots are not real.

Q.2. Find the value of k for which the quadratic equation kx (x - 2) + 6 = 0 has two equal roots.    [CBSE 2019]
Ans:
Given equation kx2-2kx + 6 = 0
For two equal roots
D = 0
⇒ b2 - 4ac = 0
⇒ 4k2 - 4k x 6 = 0
⇒ 4k2 = 24k ⇒ 4k2 = 24k = 0 ⇒  4k (k-6) = Q ⇒ k = 6
[k ≠ 0, as if k - 0 then the given equation r is not a valid equation.]

Q.3. Write the discriminant of the quadratic equation: (x + 5)2 = 2(5x - 3).

[CBSE 2019]
Ans: Given equation: (x + 5)2 - 2(5x - 3)
x2 + 25 + 10x = 10x - 6
⇒ x2 + 31 = 0
⇒ Here a = 1, b = 0, c = 31
D = b2 -4ac
= - 4 x 1 x 31 = -124
∴ Discriminant = -124

Q.4. Find the value of k, for which x = 2 is a solution of the equation kx2 + 2x - 3 = 0    [CBSE 2019]
Ans: If x = 2 is a solution of kx2 + 2x - 3 = 0
Then, k(22) + 2(2) - 3 = 0
⇒ 4k + 4 - 3 = 0
⇒ 4k = - 1 ⇒

Q.5. Using completing the square method, show that the equation x2 - 8x + 18 = 0 has no solution .    [CBSE 2019]
Ans: x- 8x + 18 = 0
⇒ x2 - 8x + 16 - 16 + 18 = 0
⇒ (x-4)2 + 2 = 0
⇒ (x - 4)2 = -2
But square of a number can’t be negative.
Hence the equation has no solution.

Q.6. Write all the values of p for which the quadratic equation x2 + px + 16 = 0 has equal roots. Find the roots of the equation so obtained.    [CBSE, Allahabad 2019]
Ans: x2 + px + 16 = 0
∴ D =b2 - 4ac
⇒ D = p2 - 4 x 1 x 16
⇒ D = p2 - 64
For equal roots, D = 0
⇒ p2 - 64 - 0
⇒  p = ± 8
For p = 8, we have
x2 4- 8x + 16=0
x2 + 4x + 4x + 16=0
x(x + 4) + 4(x + 4) =0
x = - 4 , - 4
For p = - 8, we have
x2 - 8x + 16=0
x2 - 4x - 4x + 16=0
x(x - 4) - 4(x - 4) = 0
x = 4, 4

Q.7. For what value of k, the roots of the equation x2 + 4x + k = 0 are real?

[Delhi 2019]
Ans: D = b2 - 4ac
⇒ D - (4)2 - 4 x 1 x k
⇒  D = 16 - 4k
For real roots, D > 0
⇒ 16 - 46 > 0
⇒ 16 > 46 ⇒ k < 4

Q.8. 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.    [CBSE 2018]
Ans: Let the speed be x km/h to cover a distance of 1500 km
30 minutes = 30/60 hours
= 1/2 hours

Therefore, the usual speed of the aeroplane = 500 km/h

Q.9. If x = 3 is one root of the quadratic equation x2 - 2kx - 6 = 0, then find the value of k.    [CBSE 2018]
Ans: Putting x = 3 in x2 - 2kx - 6 = 0, we get
32 - 2k x 3 - 6 = 0
⇒ 9 - 6k — 6 = 0
⇒ 3 - 6k = 0
⇒ 3 = 6k

Q.10. Divide 27 into two parts such that the sum of their reciprocals is 3/20.     [CBSE 2018 (C)
Ans:  Let one part be x and another part be 27 - x
So,

180 =x(27 - x)
x2 - 27 x + 180 = 0
(x - 15) (x - 1 2) = 0
⇒ x = 15, x = 12
So, two parts are 15 and 12

Q.11. If the roots of the quadratic equation (a - b) x+ (b - c) x + (c - a) = 0 are equal, prove that 2a = b + c.    [CBSE(F) 2017]
Ans:  Since the equation (a - b) x2 + (b - c) x + (c - a) = 0 has equal roots, therefore discriminant

Hence Proved

Q.12. Find the value of p, for which one root of the quadratic equation px2 - 14x + 8 = 0 is 6 times the other.    [CBSE (AI) 2017]
Ans: Let the roots of the given equation be α and 6α.
Thus the quadratic equation is  (x - α) (x - 6α) = 0
⇒ x2 - 7αx + 6α2 = 0 ...(i)
Given equation can be written as
...(ii)
Comparing the co-efficients in (i) & (ii)

and

Q.13. Find the roots of the quadratic equation  √2 x2 + 7x + 5 √2 = 0.

[CBSE Delhi 2017]
Ans: The given quadratic equation is
√2 x2 + 7x + 5 √2 = 0
By applying mid term splitting, we get

Q.14. If the roots of the equation (c2 - ab)x2 -2 (a2 - bc)x + b2 - ac = 0 in x are equal, then show that either a = 0 or a3 +   b3 + c3 = 3abc.    [CBSE (AI) 2017]
Ans: For equal roots D = 0
Therefore 4 (a2 - bc)2 - 4 (c2 - ab) (b2 - ac) = 0

Q.15. If the roots of the quadratic equation
(x - a) (x - b) + (x - b) (x - c) + (x - c) (x - a ) = 0
are equal, then show that a = b = c.    [CBSE (F) 2017]

Ans:

Q.16. Solve for x:
[Delhi 2017 (C)]

Ans:

x = 5 or x = -1

Q.17. Find the value of k for which the equation x2 + k(2x + k - 1) = 0 has real and equal roots.    [Delhi 2017]
Ans:
x2 + k(2x + k - 1) = 0
⇒ x2 + 2 kx +k2 - k = 0
Here, a = 1, b = 2k, c = k2 - k
Since, the roots are real and equal
∴ D = b2 - 4ac = 0
⇒ (2k)2 - 4 x 1 x (k2 - k) =0
⇒ 4k2 — 4k2 + 4k = 0
⇒ 4k = 0

Q.18. Find the value of p , for which one root of the quadratic equation px2 - 14x + 8 = 0 is 6 times the other.    [AI2017]
Ans: The given quadratic equation is
px2 - 14x + 8 = 0
Here a = p, b = - 14 and c = 8
Let α, β be the two roots of this equation such that
α = 6β
Now, sum of roots =

Also product of roots = c/a
⇒
⇒
...(ii)
Solving equations (i) and (ii) to eliminate β we get

⇒  p2 - 3p = 0
⇒ p(p - 3) = 0 ⇒  p = 0, 3
If p = 0, then the given equation will not remain quadratic.
So, p = 0 is rejected. Hence, the value of p is 3.

Q.19. If the equation (1 + m2)x2 + 2mcx + c2 - a2 = 0 has equal roots, then show that c2 = a2 (1 + m2)    [Delhi 2017]
Ans:  (1 + m2) x2 + 2mcx + (c2 - a2) = 0
Here,
a = (1 + m2)
b = 2mc
c = (c- a2)
It is given that the roots of the equation are equal; therefore, we have:

Q.20. If ad ≠ bc, then prove that the equation (a2 + b2) x2 + 2(ac + bd)x + (c2 + d2) = 0 has no real roots.    [ CBSE 2017]
Ans:
The given equation is

To find the nature of its roots, we find the discriminant.
D2 = B2 - 4AC

It is given that ad ≠ bc
⇒ ad - bc ≠ 0
⇒ (ad - bc)2 > 0 for all values of a, b, c and d
⇒ -4 (ad - bc )2 < 0
⇒ D < 0
Hence, the given equation has no real roots.

Q.21. Show that if the roots of the following quadratic equation are equal, then ad = bc
x2(a2 + b2) + 2(ac + bd)x + (c2 + d2) = 0  [Delhi 2017 (C)]
Ans: Given equation is

Q.22. If -5 is a root of the quadratic equation 2x2 + px - 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, then find the value of k.

[CBSE (AI) 2016]
Ans: Since - 5 is a root of the equation 2x2 + px - 15 = 0
∴ 2(-5)2 + p(-5) - 15 = 0
⇒ 50 - 5p - 15 = 0 or 5p = 35 or p = 7
Again p(x2 + x) 4- k = 0 or 7x2 + 7x + k = 0 has equal roots.
∴ D = 0
i.e., b2 - 4ac = 0     or 49 - 4 x 7k = 0

Q.23. If x = 2/3 and x = -3 are roots of the quadratic equation ax2 + 7x + b = 0, find the values of a and b.    [CBSE Delhi 2016]
Ans: Let us assume the quadratic equation be Ax2 + Bx + C = 0.
Sum of the roots =

Product of the roots = C/A

Q.24. A two digit number is four times the sum of the digits. It is also equal to 3 times the product of digits. Find the number.    [CBSE (F) 2016]
Ans: Let the ten’s digit be x and unit’s digit = y
Number = 10x + y
∴ 10x + y = 4(x + y)  ⇒ 6x = 3y   ⇒  2x = y   ...(i)
Again 10x + y = 3xy
10x + 2x — 3x (2x) ⇒ 12x = 6x[From equation (i)]
⇒ 6x2 - 12x = 0 ⇒ 6x(x - 2) = 0 ⇒ x = 2 (rejecting x = 0)
From (i), 2x = y ⇒ y = 4
∴ The required number is 24

Q.25. Solve for x :     [CBSE(AI) 2016]
Ans:

Q.26. Find the positive value of A, for which the equation x2 + kx + 64 = 0 and x2 - 8x + k = 0 will both have real roots.    [Foreign 2016]
Ans:  If the equation x2 + kx + 64 = 0 has real roots, then D > 0.
⇒ k2 - 4 x 64 > 0
⇒ k2 >256
⇒ k2 > (16)2
⇒ k > 16 [∵ k > 0] ...(i)
If the equation x2 - 8x + k = 0 has real roots, then
D > 0
⇒ 64 - 4k > 0
⇒ 4k < 64
⇒ k < 16  ...(ii)
From (i) and (ii), we get k = 16.

Q.27. Solve for x: x2 + 5x - (a2 + a - 6) = 0    [CBSE (F) 2015]
Ans:

## Long Answer Type Questions

Q.1. The 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.
OR
A motorboat whose speed in still water is 9 km/h, goes 15km downstream and comes back to the same spot, in a total time of 3 hours 45 minutes. Find the speed of the stream.    [CBSE 2019]

Ans:  Let the speed of the stream be x km/h.
∴ Speed of boat upstream = (15 -x) km/h.
Speed of boat downstream = (15 4- x) km/h.
According to the question,

OR
The solution is similar to the above question as only values are changed. Ans: 3 km/h

Q.2. A train travels 360 km at a uniform speed. If the speed has been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.    [CBSE 2019]
Ans: Let the uniform speed of the train be x km/h.
Then, time taken to cover 360 km
Now, new increased speed = (x 4 + 5) km/h
So, time taken to cover 360 km =
According to question,

⇒ x2 + 5x = 1800
⇒  x2 + 5x - 1800 = 0
which is a quadratic equation in x.
It can be solved by factorisation method.
⇒ x2 + 45x – 40x – 1800 = 0
⇒ x (x + 45) – 40 (x + 45) = 0
⇒ (x + 45) (x – 40) = 0
⇒ x + 45 = 0 or x – 40 =0
⇒ x = – 45 or x = 40
But speed of the train cannot be –ve.
∴ x = 40 Hence, speed of the train = 40 km/hr.

Q.3. 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 the total journey, what is the original average speed?    [CBSE 2018]
Ans: Let the original average speed of the train be x km/h
New average speed be (x + 6) km/h
Time taken for a distance of 63 km
Time taken for a distance of 72 km =

= 3(45x+ 126)
⇒ x2 + 6x = 45x +126

Therefore, the original average speed of the train is 42 km/h.

Q.4. A train covers a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km/hour, it takes 2 hours less in the journey. Find the original speed of the train.    [CBSE 2018]
Ans: Total distance travelled = 300 km
Let the speed of train = x km/hr
We know that,

Hence, time taken by train = 300/x
According to the question,
Speed of the train is increased by 5km an hour
∴ the new speed of the train = (x + 5)km/hr
Time taken to cover 300km = 300/(x + 5)
Given that time taken is 2hrs less from the previous time

⇒ 300x + 1500 – 300x = 2x (x + 5)
⇒ 1500 = 2x2 + 10x
⇒ 750 = x2 + 5x
⇒ x2 + 5x – 750 = 0
⇒ x2 + 30x – 25x – 750 = 0
⇒ x (x + 30) – 25 (x + 30) = 0
⇒ (x – 25) (x + 30) = 0
⇒ (x – 25) = 0 or (x + 30) = 0
∴ x = 25 or x = – 30
Since, speed can’t be negative.
Hence, the speed of the train is 25km/hr

Q.5. A motorboat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.    [CBSE 2018]
Ans: Let speed of stream be x km/h
Speed of motor boat in still water =18 km/h
Upstream speed = (18 - x)km/h
Downstream speed = (18 + x) km/h
Distance = 24 km
Time taken for upstream = hours
Time taken for downstream = hours

x = - 54 (rejected)  ∴ x = 6
∴ Speed of stream = 6 km/h

Q.6. Two taps running together can fill a tank in hours. If one tap takes 3 hours more than the other to fill the tank, then how much time will each tap take to fill the tank?    [AI2017]
Ans: Suppose the faster tap takes x hours to fill the tank. Therefore, the slower tap will take (x + 3) hours to fill the tank
Portion of the tank filled by the faster tap =
⇒ Portion of the tank filled by this tap in

Similarly,
Portion of tank filled by the slower tap in

It is given that the tank is filled in

⇒ 40(2x + 3) = 13x(x + 3)
⇒ 80x + 120 = 13x2 + 39x

is rejected because time cannot be negative.
Hence, the faster tap fills the cistern in 5 hours and the slower tap fills in 8 hours.

Q.7. Ram takes 6 days less than Bhagat to finish a piece of work. If both of them together can finish the work in 4 days, in how many days Bhagat alone can finish the work.    [CBSE Delhi 2017(C)]
Ans:
Let Bhagat alone can do the work in x number of days
∴ Ram takes (x - 6) number of days
Work done by Bhagat in 1 day = 1/x
Work done by Ram in 1 day =
According to the question,

Q.8. Solve for x:
[AI 2017]

Ans: The given equation is:

Q.9. Find two consecutive odd natural numbers, the sum of whose squares is 394.    [Delhi 2017 (C)]
Ans:  Let first odd natural number be (2x - 1)
and other consecutive odd natural number is (2x + 1) so,
(2x - l )2 + (2x + l )2 = 394
4x2 + 1 - 4x + 4x2 + 1 + 4x = 394
8x2 + 2 = 394
8x2 = 392
x= 49
x = 7
∴ First odd natural number = (2 x 7 - 1) = 13
Next odd natural num ber = (2 x 7 + 1 ) = 15

Q.10. The sum of the squares of two consecutive odd numbers is 394. Find the numbers.    [CBSE Delhi 2017 (C)]
Ans: Let’s assume the consecutive odd positive integer to be 2x – 1 and 2x + 1 respectively. [Keeping the common difference as 2]
Now, it’s given that the sum of their squares is 394. Which means,
(2x – 1)2 + (2x + 1)2 = 394
4x2 +1 – 4x + 4x2 +1 + 4x = 394
By cancelling out the equal and opposite terms, we get
8x2 + 2 = 394
8x2 = 392
x2 = 49
x = 7 and – 7
Since we need only consecutive odd positive integers, we only consider x = 7.
Now,
2x – 1 = 14 -1 = 13
2x + 1 = 14 + 1 = 15
Thus, the two consecutive odd positive numbers are 13 and 15 respectively.

Q.11. Solve for x:    [CBSE (AI) 2016]
Ans:

Q.12.  Find the positive value(s) of k for which both quadratic equations x2 + kx + 64 = 0 and x2 - 8x + k = 0 w ill have real roots.    [CBSE (F) 2016]
Ans: Given, the equations x2 + kx + 64 = 0 and x2 - 8x + k = 0 have real roots,
Let D1 be the discriminant of the equations x2 + kx + 64 = 0.
⇒ D1 = k2 − 4(1)(64)
⇒ D1 = k2 − 256
D2 be the discriminant of the given equations x2 - 8x + k = 0
⇒ D2 = 64 - 4(k)(1)
⇒ D2 = 64 - 4k
Given, the equations x2 + kx + 64 = 0 and x2 - 8x + k = 0 have real roots if D1, D2 ≥ 0
⇒ k2 − 256 ≥ 0 and 64 - 4k ≥ 0
⇒ (k - 16)( k + 16) ≥ 0 and 64 ≥  4k
⇒ k ≥ 16 or k ≥ - 16 and k ≤16
But k ≥ - 16 is not posisble
⇒ k ≥ 16 and k ≤16
⇒ k = 16
Hence, if the equations x2 + kx + 64 = 0 and x2 - 8x + k = 0 have real roots, then the value of k is 16.

The document Previous Year Questions: Quadratic Equations | Mathematics (Maths) Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

## FAQs on Previous Year Questions: Quadratic Equations - Mathematics (Maths) Class 10

 1. What is a quadratic equation?
Ans. A quadratic equation is a second-degree equation of the form ax²+bx+c=0 where a, b, and c are constants and x is the variable. The solutions of a quadratic equation are given by the quadratic formula, which is (-b±√(b²-4ac))/2a. Quadratic equations are used to model various real-world problems and have important applications in fields such as engineering, physics, and finance.
 2. How do you solve a quadratic equation?
Ans. To solve a quadratic equation, we can use various methods such as factoring, completing the square, or using the quadratic formula. Factoring involves finding two numbers that multiply to give the constant term and add up to give the coefficient of the linear term. Completing the square involves adding and subtracting a constant term to make a perfect square trinomial, which can then be factored. The quadratic formula can be used to find the solutions of any quadratic equation, regardless of whether it can be factored or not.
 3. What are the roots of a quadratic equation?
Ans. The roots of a quadratic equation are the values of x that satisfy the equation. Geometrically, the roots are the x-coordinates of the points where the quadratic function intersects the x-axis. If the discriminant (b²-4ac) is positive, the quadratic equation has two real roots. If the discriminant is zero, the equation has one real root of multiplicity 2. If the discriminant is negative, the equation has two complex conjugate roots.
 4. What are the applications of quadratic equations in physics?
Ans. Quadratic equations have various applications in physics. For example, they can be used to model the motion of objects under the influence of gravity, such as projectiles. The vertical motion of a projectile can be modeled using a quadratic equation, where the height of the object is given by h(t) = -16t² + vt + h₀, where t is the time, v is the initial velocity, and h₀ is the initial height. Quadratic equations can also be used to model the behavior of springs and other systems that exhibit harmonic motion.
 5. What is the discriminant of a quadratic equation?
Ans. The discriminant of a quadratic equation is the expression b²-4ac, which appears under the square root in the quadratic formula. The discriminant is a number that can be used to determine the nature of the roots of the equation. If the discriminant is positive, the equation has two real roots. If the discriminant is zero, the equation has one real root of multiplicity 2. If the discriminant is negative, the equation has two complex conjugate roots. The discriminant can also be used to find the sum and product of the roots of the equation.

## Mathematics (Maths) Class 10

90 videos|472 docs|129 tests

## Mathematics (Maths) Class 10

90 videos|472 docs|129 tests

### Top Courses for Class 10

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Track your progress, build streaks, highlight & save important lessons and more!
 (Scan QR code)
Related Searches

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;