Question

What number must be added to x^2+6x+7 to form a perfect square

Answer 1

Let the number be k,

Then adding to the expression we have,

${x}^{2} + 6x + 7 + k$
For this to be a perfect square, the discriminant must be zero.

We have,

$a=1, b=6, c=7+k$


${b}^{2} - 4ac = 0$

This implies that

${6}^{2} - 4(1)(7 +k ) = 0$
$36 - 28 - 4k = 0$
$8 - 4k = 0$
$8 = 4k$
$k = 2$
hence the number to be added is 2