A. 5, -1

B. 5, 1

C. -5, -1

D. -5, 1

3

3

3

A. 12

B. 10

C. -8

D. 4

From the formula, sum of the roots = -28/8 = −7/2 (i)

If one root is α, the other root is 6α and the sum of roots will be 7α

7α =7/2 ⇒ α = 1/2. Other root will be ½ × 6 = 3

Now the product of the roots will be =

Product of roots α × 6α = 6α

⇒ Now 6/4 =

A. -5

B. 9

C. 2

D. 11

Since it has equal roots, discriminant should be equal to zero.

(- 24)

⇒ m = 576/64 = 9.

A. p = 1

B. p = 1 or 0 or -1/2

C. p = –2

D. p = –2 or 0

⇒

⇒ 2

⇒ 2

when we take 2

Or when we take 2

2

Hence, there can be three values for

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A. cx^{2 } + bx + a =0

B. cx^{2 } + bx + c =0

C. cx^{2 } - bx + c = 0

D. cx^{2 } - bx - a = 0

Product of roots = x

Required sum of roots = (1/x

Product of roots = (1/x

Hence the required equation is x

Therefore, cx

A. b_{2} – 2c > 0 2

B. b_{2} – 2c is square of an integer & b is an integer

C. b & c are integers

D. b & c are even integers

x

Cancelling 2 from the numerator and denominator, we get -b ±√(b

Now both of the roots have to be integers and that is only possible if b happens to be an integer and

b

A. x^{2 } + x + 1 = 0

B. x^{2 } – x + 1 = 0

C. x^{2 } – x – 1 = 0

D. x^{2 } + x – 1 = 0

For the given equation the product of the roots i.e. αβ is 1. Now for the second equation, the roots are given to be 1/α and 1/β.

The sum of the roots for the equation to be calculated is 1/α + 1/β = (β + α)/αβ.

Both of these values are already calculated i.e. – 1/1 = - 1. The product of the roots for the second equation is 1/α × 1/β = 1/αβ, which can be calculated as 1/1 = 1.

The new equation is

A. 20

B. 18

C. 17

D. 19

Given A as 1, what you need to calculate is for what values of (B, C) the value of B

Take B as 6, all the given six values C can take.

Now take B as 5, C can take all the 6 values again.

Take B as 4, C can take the values from 1 to 4 i.e. 4 values.

Take B as 3, C can take only 1 and 2. Take B as 2, C can take the value 1 only.

Taking B as 1, no value of C makes it possible.

Thus the total number of ways becomes:

6 + 6 + 4 + 2 + 1 = 19 values

A. x^{2 } + 5x - 23 = 0

B. x^{2 } - 5x + 23 = 0

C. x^{2 } + 5x + 23 = 0

D. x^{2 } - 5x - 23 = 0

Equation whose roots are (3a+1), (3b+1) is

i.e

⇒