## Algebra 1: Common Core (15th Edition)

To complete the square, we will add $\dfrac{49}{4}$ on both sides and we get the approximated values of $x$ are: $x \approx 8.11,-1.11$. This method is much easier than both factorization and graphing.
$x^2-7x-9=0$ or, $x^2-7x=9$ Compare it with the standard form of quadratic equation $ax^2+bx+c$, we have $a=1, b=-7$ To complete the square, add $\dfrac{49}{4}$ on both sides. $x^2-7x+\dfrac{49}{4}=9+\dfrac{49}{4}$ $\implies (x-\dfrac{7}{2})^2=\dfrac{85}{4}$ $\implies (x-3.5)^2=21.25$ $\implies (x-3.5)=4.61$ and $\implies (x-3.5)=-4.61$ or, $x \approx 8.11,-1.11$ To complete the square, we will add $\dfrac{49}{4}$ on both sides and we get the approximated values of $x$ are: $x \approx 8.11,-1.11$. This method is much easier than both factorization and graphing since the roots are not whole numbers.