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: x² - 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) x² + (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 = (c² - 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 = 6x² [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
x² = 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.