Question

no of non -negative integral values of 'k' for which roots of the equation x2 + 6x + k = 0 are rational is-
(A) 1
(B) 2
(C) 3
(D) 4​

Answer 1

Answer:

integral values for is 1

Answer 2

Answer:

(B). 2 values of k. (k = 5, 8)

Step-by-step explanation:

In the given equation,

$x^{2}+6x+k=0$

For the roots to be rational the condition is given by,

The value in the Discriminant of the equation should be a perfect square of a number,

So,

We know that,

$Discriminant,D=\sqrt{b^{2}-4ac}$

where, in the given equation

a = 1

b = 6

and,

c = k

On comparing with the general equation $ax^{2}+bx+c=0$.

Now,

$D=\sqrt{(6)^{2}-4(1)(k)} =\sqrt{36-4k}$

Now, the value of term (36 - 4k) should be a perfect square.

For this to be possible,

at, k = 5, (36 - 4k) = 16 which is a perfect square.

at, k = 8, (36 - 4k) = 4 which is also a perfect square.

Therefore there are two non-negative integral values of k for which roots are rational.

Which are k = 5 and 8.

Hence, the correct option is (B) 2.