Chapter A.6, Problem 106AYU

To determine

To solve: Each equation by completing the square: $x^2-6x=13$.

Answer to Problem 106AYU

$x=3+\sqrt{22}, 3-\sqrt{22}$

Explanation of Solution

Given, $x^2-6x=13$.

Since the coefficient of $x^2$ is 1, we can complete the square on the left side by adding ${\left(\frac{1}{2}\cdot 6\right)}^2=9$. Of course, in an equation, whatever we add to the left side must also be added to the right side. Add 9 to both sides.

Therefore, $x^2-6x+9=13+9$.

${\left(x-3\right)}^2=22$

$x-3=\pm \sqrt{22}$

$x-3=\sqrt{22}$, $x-3=-\sqrt{22}$

$x=3+\sqrt{22}$, $x=3-\sqrt{22}$