See also: Quadratic Functions and Quadratic Equations, Quadratic Equation Formula, Quadratic Equation Factorisation, numbers

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Given a quadratic equation of the following form,

$$a<!--\; InvisibleTimes\; -->x^2+b<!--\; InvisibleTimes\; -->x+c=0$$

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

$$a<!--\; InvisibleTimes\; -->{(x+p)}^2+q=0$$

where $p=\frac{b}{2<!--\; InvisibleTimes\; -->a}$ and $q=c-\frac{b^2}{4<!--\; InvisibleTimes\; -->a}$ .

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

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

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

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

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

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Examples:

1. Find the roots of the quadratic equation

   $$x^2-4<!--\; InvisibleTimes\; -->x+3=0$$

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

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

   $$x^2-4<!--\; InvisibleTimes\; -->x=-3$$

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

   $$\begin{array}{rl}{x}^2-4<!--\; InvisibleTimes\; -->x+4& =-3+4\\ x^2-4<!--\; InvisibleTimes\; -->x+4& =1\end{array}$$

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

   $${(x-2)}^2=1$$

   Step 5. Take the square root of both sides

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

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

   $$\begin{array}{rl}x& =2\pm 1\\ x_1& =2-1=1\\ x_2& =2+1=3\end{array}$$

2. Find the roots of the quadratic equation

   $$x^2-6<!--\; InvisibleTimes\; -->x+9=0$$

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

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

   $$x^2-6<!--\; InvisibleTimes\; -->x=-9$$

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

   $$\begin{array}{rl}{x}^2-6<!--\; InvisibleTimes\; -->x+9& =-9+9\\ x^2-6<!--\; InvisibleTimes\; -->x+9& =0\end{array}$$

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

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

   Step 5. Take the square root of both sides

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

   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

   $$2<!--\; InvisibleTimes\; -->x^2+2<!--\; InvisibleTimes\; -->x+5=0$$

   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

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

   Step 5. Take the square root of both sides

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

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

   $$\begin{array}{rl}x& =-\frac{1}{2}\pm \sqrt{-\frac{9}{4}}\\ x_1& =-\frac{1}{2}-\frac{3<!--\; InvisibleTimes\; -->i}{2}\\ x_2& =-\frac{1}{2}+\frac{3<!--\; InvisibleTimes\; -->i}{2}\end{array}$$

   This quadratic equation has 2 complex roots, as can be expected because the discriminant ( $b^2-4<!--\; InvisibleTimes\; -->a<!--\; InvisibleTimes\; -->c$ ) 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.

By Jimmy Sie

See also: Quadratic Functions and Quadratic Equations, Quadratic Equation Formula, Quadratic Equation Factorisation, numbers