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Question

Number of non-negative integral values of ′k′ for which roots of the equation x2+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 x2+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=b24ac where, in the given equation
a=1,b=6 and c=k

On comparing with the general equation ax2+bx+c=0

Now,D=624×1×k=364k

Now, the value of term 364k should be a perfect square.
At k=0,364k=360=36 is a perfect square
At k=5,364k=3620=16 is a perfect square
At k=8,364k=3632=4 is a perfect square
At k=9,364k=3636=0 is a perfect square

there are four non-negative integral values of k for which roots are rational.
k=0,5,8,9

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