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** Short Answer Type Questions**

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

** [CBSE 2019 ]Ans: ** D = b

Since D < 0

Hence, roots are not real.

Ans:

For two equal roots

D = 0

⇒ b

⇒ 4k

⇒ 4k

[k ≠ 0, as if k - 0 then the given equation r is not a valid equation.]

**[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]**

Then, k(2

⇒ 4k + 4 - 3 = 0

⇒ 4k = - 1 ⇒

⇒ x

⇒ (x-4)

⇒ (x - 4)

But square of a number can’t be negative.

Hence the equation has no solution.

∴ D =b

⇒ D = p

⇒ D = p

For equal roots, D = 0

⇒ p

⇒ p = ± 8

For p = 8, we have

x

x

x(x + 4) + 4(x + 4) =0

x = - 4 , - 4

For p = - 8, we have

x

x

x(x - 4) - 4(x - 4) = 0

x = 4, 4

**[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]**

3

⇒ 9 - 6k — 6 = 0

⇒ 3 - 6k = 0

⇒ 3 = 6k

∴

So,

180 =x(27 - x)

x

(x - 15) (x - 1 2) = 0

⇒ x = 15, x = 12

So, two parts are 15 and 12

Hence Proved

Thus the quadratic equation is (x - α) (x - 6α) = 0

⇒ x

Given equation can be written as

...(ii)

Comparing the co-efficients in (i) & (ii)

and

**[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]**

Therefore 4 (a

(x - a) (x - b) + (x - b) (x - c) + (x - c) (x - a ) = 0

are equal, then show that a = b = c. [CBSE (F) 2017]

[Delhi 2017 (C)]

x = 5 or x = -1

⇒ x

Here, a = 1, b = 2k, c = k

Since, the roots are real and equal

∴ D = b

⇒ (2k)

⇒ 4k

⇒ 4k = 0

∴

px

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

⇒ 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.

Ans:

To find the nature of its roots, we find the discriminant.

D

It is given that ad ≠ bc

⇒ ad - bc ≠ 0

⇒ (ad - bc)

⇒ -4 (ad - bc )

⇒ D < 0

Hence, the given equation has no real roots.

**[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]**

Sum of the roots =

⇒

Product of the roots = C/A

⇒

Number = 10x + y

∴ 10x + y = 4(x + y) ⇒ 6x = 3y ⇒ 2x = y ...(i)

Again 10x + y = 3xy

10x + 2x — 3x (2x) ⇒ 12x = 6x

⇒ 6x

From (i), 2x = y ⇒ y = 4

∴ The required number is 24

⇒ k

⇒ k

⇒ k

⇒ k

If the equation x

D

⇒ 64 - 4k

⇒ 4k

⇒ k

From (i) and (ii), we get k = 16.

**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.ORA 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]**

∴ 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

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,

But x cannot be negative, so x ≠ - 45

Therefore, x = 40

Hence, the uniform speed of train is 40 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

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

Therefore, time taken to cover 300 km = ...(i)

When its speed is increased by 5 km/h, then the time taken by the train to cover the distance of ...(ii)

According to the question, hours = 2 hours

Therefore, the usual speed of the train = 25 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

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

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.

Ans:

∴ 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,

[AI 2017]

and other consecutive odd natural number is (2x + 1) so,

(2x - l )

4x

8x

8x

x

x = 7

∴ First odd natural number = (2 x 7 - 1) = 13

Next odd natural num ber = (2 x 7 + 1 ) = 15

Hence, the numbers are 13 and 15 or -15 and -13.

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