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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
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B
2
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C
3
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D
4
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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 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.
$$\therefore\,k=0,5,8,9$$

Mathematics

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