## Intermediate Algebra for College Students (7th Edition)

$x=1+\sqrt{7}$
Set up the equation: Let $x$ be the number. Note that $x>0$. $x^2 - (6+2x)=0$ $x^2 - 6 - 2x =0$ $x^2 -2x-6=0$ $x^2 -2x=6$ Compute for $x$ by completing the square. The coefficient of the $x$-term is $-2$; $\frac{-2}{2}=-1$; $-1^2=1$ Add $1$ to both sides to complete the square. $x^2 -2x+1=6+1$ $x^2 -2x+1=7$ $(x-1)^2=7$ $x-1=±\sqrt{7}$ $x=1±\sqrt{7}$ Since $x>0$, thus, $x=1+\sqrt{7}$.