Answer
-8 or 8
Work Step by Step
Let the unknown number be $x$. This means that one-half of the number is $\frac{x}{2}$.
As a result, the square of the unknown number is $x^{2}$ whereas the square of one-half of the number is $(\frac{x}{2})^{2}$.
Since the sum of the square of a number and the square of
one-half of the number is 80, we write the following equation and solve it:
$x^{2}+(\frac{x}{2})^{2}=80$
$x^{2}+(\frac{x^{2}}{4})=80$
$\frac{4x^{2}+x^{2}}{4}=80$
$\frac{5x^{2}}{4}=80$
$5x^{2}=320$
$x^{2}=\frac{320}{5}$
$x^{2}=64$
$x=\pm\sqrt {64}$
$x=\pm 8$
Check when x=8:
$8^{2}+(\frac{8}{2})^{2}=80$
$8^{2}+(4)^{2}=80$
$64+16=80$
$80=80$
Check when x=-8:
$(-8)^{2}+(\frac{-8}{2})^{2}=80$
$(-8)^{2}+(-4)^{2}=80$
$64+16=80$
$80=80$
Therefore, the unknown number can be either -8 or 8.
