Given a quadratic equation of the following form,

$a{x}^{2}+bx+c=0$

we can find the roots by first converting the above form into the one below:

$a{\left(x+p\right)}^{2}+q=0$

where $p=\frac{b}{2a}$ and $q=c-\frac{{b}^{2}}{4a}$ .

The idea of completing the square comes from the fact that if we have a quadratic equation of the form:

${\left(x+m\right)}^{2}=n$

then it's easy to solve by taking the square root of both sides.

$\begin{array}{rl}\sqrt{{\left(x+m\right)}^{2}}& =±\sqrt{n}\\ x+m& =±\sqrt{n}\\ x& =-m±\sqrt{n}\end{array}$

Follow the steps below to solve a quadratic equation in the general form by completing the square.

 Original equation $a{x}^{2}+bx+c=0$ Step 1. Divide the equation by $a$ to get the coefficient of ${x}^{2}$ equals to $1$ ${x}^{2}+\frac{bx}{a}+\frac{c}{a}=0$ Step 2. Move the constant term to the right hand side ${x}^{2}+\frac{bx}{a}=-\frac{c}{a}$ Step 3. Add ${\left(\frac{b}{2a}\right)}^{2}$ to both sides of the equation ${x}^{2}+\frac{bx}{a}+{\left(\frac{b}{2a}\right)}^{2}=-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}$ Step 4. We can now re-write the left hand side as a complete square ${\left(x+\frac{b}{2a}\right)}^{2}=-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}$ Step 5. Take the square root of both sides $\begin{array}{rl}\sqrt{{\left(x+\frac{b}{2a}\right)}^{2}}& =±\sqrt{-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}}\\ x+\frac{b}{2a}& =±\sqrt{-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}}\end{array}$ Step 6. Move the constant term from left hand side to the right hand side, then solve for $x$ $x=-\frac{b}{2a}±\sqrt{-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}}$

Examples:

1. Find the roots of the quadratic equation

${x}^{2}-4x+3=0$

Step 1. We can skip this step since in this case $a=1$

Step 2. Move the constant term to the right hand side

${x}^{2}-4x=-3$

Step 3. Add ${\left(\frac{4}{2}\right)}^{2}$ to both sides of the equation (i.e. add $4$)

$\begin{array}{rl}{x}^{2}-4x+4& =-3+4\\ {x}^{2}-4x+4& =1\end{array}$

Step 4. Re-write left hand side as a complete square

${\left(x-2\right)}^{2}=1$

Step 5. Take the square root of both sides

$\begin{array}{rl}\sqrt{{\left(x-2\right)}^{2}}& =±\sqrt{1}\\ x-2& =±1\end{array}$

Step 6. Move the constant term from left hand side to the right hand side, then solve for $x$

$\begin{array}{rl}x& =2±1\\ {x}_{1}& =2-1=1\\ {x}_{2}& =2+1=3\end{array}$
2. Find the roots of the quadratic equation

${x}^{2}-6x+9=0$

Step 1. We can skip this step since in this case $a=1$

Step 2. Move the constant term to the right hand side

${x}^{2}-6x=-9$

Step 3. Add ${\left(\frac{6}{2}\right)}^{2}$ to both sides of the equation (i.e. add $9$)

$\begin{array}{rl}{x}^{2}-6x+9& =-9+9\\ {x}^{2}-6x+9& =0\end{array}$

Step 4. Re-write left hand side as a complete square

${\left(x-3\right)}^{2}=0$

Step 5. Take the square root of both sides

$\begin{array}{rl}\sqrt{{\left(x-3\right)}^{2}}& =±\sqrt{0}\\ x-3& =0\end{array}$

Step 6. Move the constant term from left hand side to the right hand side, then solve for $x$

$x=3$

In this case, we could have skipped steps 2 & 3 too, because the original equation is already a complete square, i.e. we can straight away re-write the original equation as a complete square as shown in step 4.

3. Find the roots of the quadratic equation

$2{x}^{2}+2x+5=0$

Step 1. Divide the equation by $2$

${x}^{2}+x+\frac{5}{2}=0$

Step 2. Move the constant term to the right hand side

${x}^{2}+x=-\frac{5}{2}$

Step 3. Add ${\left(\frac{1}{2}\right)}^{2}$ to both sides of the equation (i.e. add $\frac{1}{4}$)

$\begin{array}{rl}{x}^{2}+x+\frac{1}{4}& =-\frac{5}{2}+\frac{1}{4}\\ {x}^{2}+x+\frac{1}{4}& =-\frac{9}{4}\end{array}$

Step 4. Re-write left hand side as a complete square

${\left(x+\frac{1}{2}\right)}^{2}=-\frac{9}{4}$

Step 5. Take the square root of both sides

$\begin{array}{rl}\sqrt{{\left(x+\frac{1}{2}\right)}^{2}}& =±\sqrt{-\frac{9}{4}}\\ x+\frac{1}{2}& =±\sqrt{-\frac{9}{4}}\end{array}$

Step 6. Move the constant term from left hand side to the right hand side, then solve for $x$

$\begin{array}{rl}x& =-\frac{1}{2}±\sqrt{-\frac{9}{4}}\\ {x}_{1}& =-\frac{1}{2}-\frac{3i}{2}\\ {x}_{2}& =-\frac{1}{2}+\frac{3i}{2}\end{array}$

This quadratic equation has 2 complex roots, as can be expected because the discriminant ( ${b}^{2}-4ac$ ) is negative.

Confused and have questions? We’ve got answers. With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field.