Question

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

A
1
B
2
C
3
D
4

Solution

## 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 squareAt $$k=5, \,36-4k=36-20=16$$ is a perfect squareAt $$k=8, \,36-4k=36-32=4$$ is a perfect squareAt $$k=9, \,36-4k=36-36=0$$ is a perfect square$$\therefore$$ there are four non-negative integral values of $$k$$ for which roots are rational.$$\therefore\,k=0,5,8,9$$Mathematics

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