## Precalculus (6th Edition) Blitzer

The roots of the equation ${{y}^{2}}-6y-4=0$ by the use of completing the square method are: $y=\left\{ 3+\sqrt{13}\text{ ,}3-\sqrt{13} \right\}$.
In completing the square method, convert the equation into the form of ${{\left( a+b \right)}^{2}}$ or ${{\left( a-b \right)}^{2}}$. We have: ${{y}^{2}}-6y-4=0$ ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ But $a=y$, and \begin{align} & 2ab=6y \\ & 2b=6 \\ & b=3 \end{align} So, ${{b}^{2}}={{3}^{2}}=9$ Now, add 9 to both sides of ${{y}^{2}}-6y-4=0$, \begin{align} & {{y}^{2}}-6y-4+9=9 \\ & {{y}^{2}}-6y+9=9+4 \\ & {{y}^{2}}-6y+9=13 \end{align} But, the equation ${{y}^{2}}-6y+9$ is the complete square of ${{\left( y-3 \right)}^{2}}$, ${{\left( y-3 \right)}^{2}}=13$ Taking the square root on both sides, $y-3=\pm \sqrt{13}$ Therefore, the roots are $y=3+\sqrt{13}\text{ or }y=3-\sqrt{13}$.