Question

Complete the square on the following quadratic

x^2+16x+9=0

After removing C, what value is added to both sides of the equation?

Answer 1

Answer:

Completing the square we have

$x_{1} =\sqrt{55}-8\\\\x_{2}=-\sqrt{55}-8$

Step-by-step explanation:

$x^{2} +16x+9=0$

we have to divide the coefficient of x by half, and then raise it to the square, that number is added and subtracted at the same time

$x^{2}+16x+64-64+9=0$

$(x+8)^{2} -55=0$

$(x+8)^{2} =55$

later we can write

$\sqrt{(x+8)^{2}}= \sqrt{55}$

we know

$\sqrt[2]{a^{2}} =abs(a)$

so we have

abs(x+8)=$\sqrt{55}$

we have

$x_{1} =$x+8=$\sqrt{55}$

$x_{2}=$x+8=$-\sqrt{55}$

finally we have

$x_{1}=\sqrt{55}-8$

$x_{2} = -\sqrt{55}-8$