How do you find the value of a such that $y^{2} + 12 y + a$ $y^2 +12y + a$ is a perfect square?

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Answer 1 · 2015-04-07T17:10:11

${(y + n)}^{2} = y^{2} + 2 n y + n^{2}$ $(y+n)^2 = y^2+2ny+n^2$

If the coefficient of $y$ $y$ in a perfect square is $12$ $12$
then
$2 n = 12$ $2n = 12$
$\to n = 6$ $rarr n= 6$
and
$n^{2} = 36$ $n^2 = 36$

$y^{2} + 12 y + 36 = {(y + 6)}^{2}$ $y^2+12y+36 = (y+6)^2$

$a = 36$ $a=36$
makes $y^{2} + 12 y + a$ $y^2+12y+a$ a perfect square

How do you find the value of a such that y2+12y+ay^2 +12y + a is a perfect square?