Q.1. What will be the nature of roots of quadratic equation 2x^{2}  4x + 3 = 0?
[CBSE 2019 ]
Ans: D = b^{2}  4ac = 4^{2}  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 kx^{2}2kx + 6 = 0
For two equal roots
D = 0
⇒ b^{2}  4ac = 0
⇒ 4k^{2}  4k x 6 = 0
⇒ 4k^{2} = 24k ⇒ 4k^{2} = 24k = 0 ⇒ 4k (k6) = 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)
x^{2} + 25 + 10x = 10x  6
⇒ x^{2} + 31 = 0
⇒ Here a = 1, b = 0, c = 31
D = b^{2} 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 kx^{2} + 2x  3 = 0 [CBSE 2019]
Ans: If x = 2 is a solution of kx^{2} + 2x  3 = 0
Then, k(2^{2}) + 2(2)  3 = 0
⇒ 4k + 4  3 = 0
⇒ 4k =  1 ⇒
Q.5. Using completing the square method, show that the equation x^{2}  8x + 18 = 0 has no solution . [CBSE 2019]
Ans: x^{2 } 8x + 18 = 0
⇒ x^{2}  8x + 16  16 + 18 = 0
⇒ (x4)^{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 x^{2} + px + 16 = 0 has equal roots. Find the roots of the equation so obtained. [CBSE, Allahabad 2019]
Ans: x^{2} + px + 16 = 0
∴ D =b^{2}  4ac
⇒ D = p^{2}  4 x 1 x 16
⇒ D = p^{2}  64
For equal roots, D = 0
⇒ p^{2}  64  0
⇒ p = ± 8
For p = 8, we have
x^{2} 4 8x + 16=0
x^{2} + 4x + 4x + 16=0
x(x + 4) + 4(x + 4) =0
x =  4 ,  4
For p =  8, we have
x^{2}  8x + 16=0
x^{2}  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 x^{2} + 4x + k = 0 are real?
[Delhi 2019]
Ans: D = b^{2}  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 x^{2}  2kx  6 = 0, then find the value of k. [CBSE 2018]
Ans: Putting x = 3 in x^{2}  2kx  6 = 0, we get
3^{2}  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)
x^{2}  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^{2 }+ (b  c) x + (c  a) = 0 are equal, prove that 2a = b + c. [CBSE(F) 2017]
Ans: Since the equation (a  b) x^{2} + (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 px^{2}  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
⇒ x^{2}  7αx + 6α^{2} = 0 ...(i)
Given equation can be written as
...(ii)
Comparing the coefficients in (i) & (ii)
and
Q.13. Find the roots of the quadratic equation √2 x^{2} + 7x + 5 √2 = 0.
[CBSE Delhi 2017]
Ans: The given quadratic equation is
√2 x^{2} + 7x + 5 √2 = 0
By applying mid term splitting, we get
Q.14. If the roots of the equation (c^{2}  ab)x^{2} 2 (a2  bc)x + b^{2}  ac = 0 in x are equal, then show that either a = 0 or a^{3} + b^{3} + c3 = 3abc. [CBSE (AI) 2017]
Ans: For equal roots D = 0
Therefore 4 (a^{2}  bc)^{2}  4 (c^{2}  ab) (b^{2}  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 x^{2} + k(2x + k  1) = 0 has real and equal roots. [Delhi 2017]
Ans: x^{2} + k(2x + k  1) = 0
⇒ x^{2} + 2 kx +k^{2}  k = 0
Here, a = 1, b = 2k, c = k^{2}  k
Since, the roots are real and equal
∴ D = b^{2}  4ac = 0
⇒ (2k)^{2}  4 x 1 x (k^{2}  k) =0
⇒ 4k^{2} — 4k^{2} + 4k = 0
⇒ 4k = 0
∴
Q.18. Find the value of p , for which one root of the quadratic equation px^{2}  14x + 8 = 0 is 6 times the other. [AI2017]
Ans: The given quadratic equation is
px^{2}  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
⇒ p^{2}  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 + m^{2})x^{2} + 2mcx + c^{2}  a^{2} = 0 has equal roots, then show that c^{2} = a^{2} (1 + m^{2}) [Delhi 2017]
Ans: (1 + m^{2}) x^{2} + 2mcx + (c^{2}  a^{2}) = 0
Here,
a = (1 + m^{2})
b = 2mc
c = (c^{2 } a^{2})
It is given that the roots of the equation are equal; therefore, we have:
Q.20. If ad ≠ bc, then prove that the equation (a^{2} + b^{2}) x^{2} + 2(ac + bd)x + (c^{2} + d^{2}) = 0 has no real roots. [ CBSE 2017]
Ans: The given equation is
To find the nature of its roots, we find the discriminant.
D^{2} = B^{2}  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
x^{2}(a^{2} + b^{2}) + 2(ac + bd)x + (c^{2} + d^{2}) = 0 [Delhi 2017 (C)]
Ans: Given equation is
Q.22. If 5 is a root of the quadratic equation 2x^{2} + px  15 = 0 and the quadratic equation p(x^{2} + x) + k = 0 has equal roots, then find the value of k.
[CBSE (AI) 2016]
Ans: Since  5 is a root of the equation 2x^{2} + px  15 = 0
∴ 2(5)^{2} + p(5)  15 = 0
⇒ 50  5p  15 = 0 or 5p = 35 or p = 7
Again p(x^{2} + x) 4 k = 0 or 7x^{2} + 7x + k = 0 has equal roots.
∴ D = 0
i.e., b^{2}  4ac = 0 or 49  4 x 7k = 0
⇒
Q.23. If x = 2/3 and x = 3 are roots of the quadratic equation ax^{2} + 7x + b = 0, find the values of a and b. [CBSE Delhi 2016]
Ans: Let us assume the quadratic equation be Ax^{2} + 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^{2 }[From equation (i)]
⇒ 6x^{2}  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 x^{2} + kx + 64 = 0 and x^{2}  8x + k = 0 will both have real roots. [Foreign 2016]
Ans: If the equation x^{2} + kx + 64 = 0 has real roots, then D > 0.
⇒ k^{2}  4 x 64 > 0
⇒ k^{2} >256
⇒ k^{2} > (16)^{2}
⇒ k > 16 [∵ k > 0] ...(i)
If the equation x^{2}  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: x^{2} + 5x  (a^{2} + a  6) = 0 [CBSE (F) 2015]
Ans:
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,
⇒ x^{2} + 5x = 1800
⇒ x^{2} + 5x  1800 = 0
which is a quadratic equation in x.
It can be solved by factorisation method.
⇒ x^{2} + 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)
⇒ x^{2} + 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 = 2x^{2} + 10x
⇒ 750 = x^{2} + 5x
⇒ x^{2} + 5x – 750 = 0
⇒ x^{2} + 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 = 13x^{2} + 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
4x^{2} + 1  4x + 4x^{2} + 1 + 4x = 394
8x^{2} + 2 = 394
8x^{2} = 392
x^{2 }= 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
4x^{2} +1 – 4x + 4x^{2} +1 + 4x = 394
By cancelling out the equal and opposite terms, we get
8x^{2} + 2 = 394
8x^{2} = 392
x^{2} = 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 x^{2} + kx + 64 = 0 and x^{2}  8x + k = 0 w ill have real roots. [CBSE (F) 2016]
Ans: Given, the equations x^{2} + kx + 64 = 0 and x^{2}  8x + k = 0 have real roots,
Let D_{1} be the discriminant of the equations x^{2} + kx + 64 = 0.
⇒ D_{1} = k^{2} − 4(1)(64)
⇒ D_{1} = k^{2} − 256
D_{2} be the discriminant of the given equations x^{2}  8x + k = 0
⇒ D_{2} = 64  4(k)(1)
⇒ D_{2} = 64  4k
Given, the equations x^{2} + kx + 64 = 0 and x^{2}  8x + k = 0 have real roots if D_{1}, D_{2} ≥ 0
⇒ k^{2} − 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 x^{2} + kx + 64 = 0 and x^{2}  8x + k = 0 have real roots, then the value of k is 16.
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