Question

Number of non-negative integral values of ${}^{\prime }{k}^{\prime }$ for which roots of the equation $x^2+6x+k=0$ are rational is-

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

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 $a{x}^2+bx+c=0$

Now,$D=\sqrt{6^2-4\times 1\times k}=\sqrt{36-4k}$

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

At $k=0,36-4k=36-0=36$ is a perfect square

At $k=5,36-4k=36-20=16$ is a perfect square

At $k=8,36-4k=36-32=4$ is a perfect square

At $k=9,36-4k=36-36=0$ is a perfect square

$\therefore$ there are four non-negative integral values of $k$ for which roots are rational.

$\therefore k=0,5,8,9$