Find the value of “k” so that the quadratic polynomial has equal zeroes.

The question is this:

Find the the value(s) of $k$ so that the quadratic polynomial $kx^2 + x + k$ has equal zeroes.

Answers along with appropriate explanations would be appreciated.

Thanks.

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If $k=0$ then the polynomial is not quadratic. Does it have equal zeroes then? – Servaes Jun 30 '14 at 10:20
The question is concerning a quadratic equation. If we change the type of equation, then it would completely defeat the purpose of the question. The question is for a quadratic equation, and not a linear one. – EuclidAteMyBreakfast Jun 30 '14 at 10:23

$\displaystyle{x}_{1,2}=\frac{-1\pm\sqrt{1-4k^2}}{2k}$

$\displaystyle\sqrt{1-4k^2}=0\implies{x}_{1}={x}_{2}=-\frac{1}{2k}$

$\displaystyle\sqrt{1-4k^2}=0\implies1-4k^2=0\implies{k}=\pm\frac{1}{2}$

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Thanks a lot, but I guess using b^2 - 4ac = 0 to get the answer is a bit simpler. :) – EuclidAteMyBreakfast Jun 30 '14 at 7:12
@EuclidAteMyBreakfast: That's exactly what I'm doing here, but it also explains (to you) why for this value of $k$ you will get the same ${x}_{1,2}$. – barak manos Jun 30 '14 at 7:14

Hint : The quadratic equation $ax^2+bx+c=0$ has equal roots if and only if $b^2-4ac=0$. You may now calculate the value(s) of $k$ accordingly.

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I loved how you didn't give me the answer right away, so that I did it myself. ;) Thanks a lot. :) – EuclidAteMyBreakfast Jun 30 '14 at 7:13
ok thanks ! Actual formula for the roots of a quadratic equation is $x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$ as said by Barak manos. – Debashish Jun 30 '14 at 7:16