Section 4.2
Completing the Square
We'll start this section with the common notion that if $x^2=9$, then $x=3$ or $x=-3$. In general, $$\text{If }\;x^2=a,\;\text{then }\;x=\pm\sqrt{a}.$$
Note:
- If $a=0$, then there is just one solution to the equation $x^2=a$.
- If $a>0$, then ther are two solutions (one positive and the other negative) to the equation $x^2=a$.
- And if $a<0$, then ther are no solutions to the equation
$x^2=a$.
If the left had side of the equation is not a perfect square, there are two techniques we can employ to improve it so that it is one.
- Factoring or
- Adding a certain quantity to both sides, completing the square.
a. \begin{align}
x^2+4x+4&=10\\
(x+2)^2&=10&&\textit{Factor the left side.}\\
x+2&=\pm\sqrt{10}\\
x&=-2\pm\sqrt{10}
\end{align}
b. \begin{align}
x^2-8x+16&=25\\
(x-4)^2&=25&&\textit{Factor the left side.}\\
x-4&=\pm\sqrt{25}\\
x&=4\pm 5\\
\text{So,}\;\;x=9\;\;\text{o}&\text{r}\;\;x=-1
\end{align}
The following examples will employ the second technique. In each case we will add a certain quantity to both sides. This quantity will always be the square of $\frac{1}{2}$ the coefficient of the $x$ term.
c.\begin{align}
x^2-6x+2&=0\\
x^2-6x\;\;\;\;\;\;&=-2&&\textit{Making room to complete the square.}\\
x^2-6x+9&=-2+9&&\textit{Adding $(\frac{1}{2}\cdot -6)^2=(-3)^2=9$ to both sides, completing the square.}\\
(x-3)^2&=7&&\textit{Factor the left side.}\\
x-3&=\pm\sqrt{7}\\
x&=3\pm\sqrt{7}
\end{align}
d. \begin{align}
x^2+12x+33&=0\\
x^2+12x\;\;\;\;\;\;\;\;&=-33&&\textit{Making room to complete the square.}\\
x^2+12x+36&=-33+36&&\textit{Adding $(\frac{1}{2}\cdot 12)^2=6^2=36$ to both sides, completing the square.}\\
(x+6)^2&=3&&\textit{Factor the left side.}\\
x+6&=\pm\sqrt{3}\\
x&=-6\pm\sqrt{3}
\end{align}
The technique employed in examples (c.) and (d.) above is known as "completing the square." It is particularly appropriate when the coefficient of $x^2$ in the quadratic equation is 1 and when the coefficient of $x$ is an even number.
Solving quadratic equations by completing the square: Solving Quadratic Equations by Completing the Square
Exercises
Solve the following quadratic equations, if possible.\begin{array}{l l}
1. \;\;{{x}^{2}}-6x+9=3\;\;& 2. \;\;{{x}^{2}}+10x+25=5\;\;& 3. \;\;{{x}^{2}}-4x+4=0\;\; \\
4. \;\;{{x}^{2}}+8x+16=13\;\;& 5. \;\;{{x}^{2}}-20x+100=64\;\;& 6. \;\;{{x}^{2}}-2x=4\;\; \\
7. \;\;{{x}^{2}}+14x=3\;\;& 8. \;\;{{x}^{2}}-4x=0\;\;& 9. \;\;{{x}^{2}}-18x=13\;\; \\
10. \;\;{{x}^{2}}=64\;\;& 11. \;\;{{x}^{2}}+6x+2=0\;\;& 12. \;\;{{x}^{2}}-4x-7=0\;\; \\
13. \;\;{{x}^{2}}-24x+3=0\;\;& 14. \;\;{{x}^{2}}-6x-7=0\;\;& 15. \;\;{{x}^{2}}-2x-8=0\;\; \\
16. \;\;{{x}^{2}}-16x-36=0\;\;& 17. \;\;{{x}^{2}}+12x+11=0\;\;& 18. \;\;{{x}^{2}}-25=0\;\; \\
19. \;\;{{x}^{2}}-6x=0\;\;& 20. \;\;{{x}^{2}}-2x+6=0\;\;& 21. \;\;(x-4)(x+10)+2=0\;\; \\
22. \;\;(x-3)(x+5)=1\;\;& 23. \;\;(x-4)(x-10)=10\;\;& 24. \;\;(2x-4)(x-3)=(x+1)(x-1)\;\; \\
25. \;\;(x-7)(x+3)=6\;\;& 26. \;\;(x-3)(x+5)=9\;\;& 27. \;\;(2x-3)(x+5)={{(x+1)}^{2}}-x\;\;
\end{array}