SOLVING QUADRATIC EQUATION : Solving by Completing the Square

Introduction

Apart from factoring method and quadratic formula, quadratic equation can be solved by using Completing the Square method.

How to solve - Solving Quadratic Equation : Solving by Completing the Square?

To solve a quadratic equation by completing the square.

1. Divide all terms with value of a.
2. Isolate the terms in x on one side of the equation.
3. Add the square of one-half the coefficient of x or ${\left(\frac{b}{2}\right)}^2$to both sides of the equation.
4. Write the square root of both sides of the resulting equation and solve for x.

Example : Solve the following by using completing the square method.

$\begin{array}{l}a)x^2-4x=-3\{makesureaispositive1\\ x^2-4x+{\left(-\frac{4}{2}\right)}^2=-3+{\left(-\frac{4}{2}\right)}^2\{add{\left(\frac{b}{2}\right)}^2={\left(\frac{-4}{2}\right)}^2=\underset{onleftofequation}{\underbrace{{\left(-2\right)}^2}}=\underset{onrightofequation}{\underbrace{4}}\\ x^2-4x+{\left(-2\right)}^2=-3+4\\ {\left(x-2\right)}^2=1\\ \sqrt{{\left(x-2\right)}^2}=\pm \sqrt{1}\\ x-2=\pm 1\\ x-2=1,x-2=-1\\ \underset{¯}{\underset{¯}{x=3,x=1}}\end{array}$

$\begin{array}{l}b)-x^2+4x+8=0\{makesureaispositive1\\ \frac{-x^2}{-1}+\frac{4x}{-1}+\frac{8}{-1}=\frac{0}{-1}\{dividealltermswith-1\\ x^2-4x-8=0\\ x^2-4x=8\\ x^2-4x+{\left(\frac{-4}{2}\right)}^2=8+{\left(\frac{-4}{2}\right)}^2\{add{\left(\frac{b}{2}\right)}^2={\left(\frac{-4}{2}\right)}^2=\underset{onleftofequation}{\underbrace{{\left(-2\right)}^2}}=\underset{onrightofequation}{\underbrace{4}}\\ x^2-4x+{\left(-2\right)}^2=8+4\\ {\left(x-2\right)}^2=12\\ \sqrt{{\left(x-2\right)}^2}=\pm \sqrt{12}\\ x-2=\pm 3.46\\ x-2=3.46,x-2=-3.46\\ \underset{¯}{\underset{¯}{x=5.46,x=-1.46}}\end{array}$

$\begin{array}{l}c)2x^2-x-1=0\\ \frac{2x^2}{2}-\frac{x}{2}-\frac{1}{2}=\frac{0}{2}\\ x^2-\frac{1}{2}x-\frac{1}{2}=0\\ x^2-\frac{1}{2}x=\frac{1}{2}\\ x^2-\frac{1}{2}x+{\left(\frac{-\frac{1}{2}}{2}\right)}^2=\frac{1}{2}+{\left(\frac{-\frac{1}{2}}{2}\right)}^2\\ x^2-4x+{\left(-\frac{1}{4}\right)}^2=\frac{8}{16}+\left(\frac{1}{16}\right)\\ {\left(x-\frac{1}{4}\right)}^2=\frac{9}{16}\\ x-\frac{1}{4}=\pm \sqrt{\frac{9}{16}}\\ x-2=\pm 3.46\\ x-2=3.46,x-2=-3.46\\ \underset{¯}{\underset{¯}{x=5.46,x=-1.46}}\end{array}$

$\begin{array}{l}d)2x^2-x-3=0\\ \frac{2x^2-x-3}{2}=\frac{0}{2}\\ x^2-\frac{x}{2}-\frac{3}{2}=0\\ x^2-\frac{1}{2}x-\frac{3}{2}=0\\ x^2-\frac{1}{2}x+{\left(\frac{-\frac{1}{2}}{2}\right)}^2=\frac{3}{2}+{\left(\frac{-\frac{1}{2}}{2}\right)}^2\\ x^2-\frac{1}{2}x+{\left(-\frac{1}{4}\right)}^2=\frac{3}{2}+{\left(-\frac{1}{4}\right)}^2\\ {\left(x-\frac{1}{4}\right)}^2=\frac{3}{2}+\frac{1}{16}\\ {\left(x-\frac{1}{4}\right)}^2=\frac{25}{16}\\ {\left(x-\frac{1}{4}\right)}^2=\frac{25}{16}\\ \sqrt{{\left(x-\frac{1}{4}\right)}^2}=\pm \sqrt{\frac{25}{16}}\\ x-\frac{1}{4}=\pm \frac{5}{4}\\ x=\frac{1}{4}\pm \frac{5}{4}\\ x=\frac{6}{4}orx=-\frac{4}{4}\\ x=\frac{3}{2}orx=-1\end{array}$

Try it yourself - Solving Quadratic Equation : Solving quadratic equation by using Completing the Square

$\begin{array}{l}a)4x^2-8x-5=0\\ b)2x^2-x-6=0\\ c)12x^2-18x=0\end{array}\}\begin{array}{c}ans-x=\frac{5}{2},-\frac{1}{2}\\ ans-x=2,-\frac{3}{2}\\ ans-x=0,\frac{3}{2}\end{array}$

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