For what value of $k$ $k$ is $x^{2} - 21 x + k$ $x^2-21x+k$ a perfect square trinomial?

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Answer 1 · 2017-03-26T08:09:21

$k = \frac{441}{4} = 110.25$ $k = 441/4 = 110.25$

Suppose:

$x^{2} - 21 x + k = {(x + e)}^{2}$ $x^2-21x+k = (x+e)^2$

$x^{2} - 21 x + k = x^{2} + 2 e x + e^{2}$ $color(white)(x^2-21x+k) = x^2+2ex+e^2$

Equating coefficients, we have:

$2 e = - 21$ $2e=-21$

So:

$e = - \frac{21}{2}$ $e = -21/2$

and

$k = e^{2} = {(- \frac{21}{2})}^{2} = \frac{441}{4} = 110.25$ $k = e^2 = (-21/2)^2 = 441/4 = 110.25$

For what value of kkk is x^2-21x+kx2−21x+kx^2-21x+k a perfect square trinomial?