I Do Maths · Quadratic Equations — Completing the Square

Given a quadratic equation of the following form,

a x^{2} + b x + c = 0

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

a {(x + p)}^{2} + q = 0

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

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{aligned} \sqrt{{(x + m)}^{2}} & = \pm \sqrt{n} \\ x + m & = \pm \sqrt{n} \\ x & = - m \pm \sqrt{n} \end{aligned}

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

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

Examples:

Find the roots of the quadratic equation
$x^{2} - 4 x + 3 = 0$
Step 1. We can skip this step since in this case $a = 1$

Step 2. Move the constant term to the right hand side
$x^{2} - 4 x = - 3$
Step 3. Add ${(\frac{4}{2})}^{2}$ to both sides of the equation (i.e. add $4$ )
$\begin{aligned} x^{2} - 4 x + 4 & = - 3 + 4 \\ x^{2} - 4 x + 4 & = 1 \end{aligned}$
Step 4. Re-write left hand side as a complete square
${(x - 2)}^{2} = 1$
Step 5. Take the square root of both sides
$\begin{aligned} \sqrt{{(x - 2)}^{2}} & = \pm \sqrt{1} \\ x - 2 & = \pm 1 \end{aligned}$
Step 6. Move the constant term from left hand side to the right hand side, then solve for $x$
$\begin{aligned} x & = 2 \pm 1 \\ x_{1} & = 2 - 1 = 1 \\ x_{2} & = 2 + 1 = 3 \end{aligned}$
Find the roots of the quadratic equation
$x^{2} - 6 x + 9 = 0$
Step 1. We can skip this step since in this case $a = 1$

Step 2. Move the constant term to the right hand side
$x^{2} - 6 x = - 9$
Step 3. Add ${(\frac{6}{2})}^{2}$ to both sides of the equation (i.e. add $9$ )
$\begin{aligned} x^{2} - 6 x + 9 & = - 9 + 9 \\ x^{2} - 6 x + 9 & = 0 \end{aligned}$
Step 4. Re-write left hand side as a complete square
${(x - 3)}^{2} = 0$
Step 5. Take the square root of both sides
$\begin{aligned} \sqrt{{(x - 3)}^{2}} & = \pm \sqrt{0} \\ x - 3 & = 0 \end{aligned}$
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.
Find the roots of the quadratic equation
$2 x^{2} + 2 x + 5 = 0$
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 ${(\frac{1}{2})}^{2}$ to both sides of the equation (i.e. add $\frac{1}{4}$ )
$\begin{aligned} x^{2} + x + \frac{1}{4} & = - \frac{5}{2} + \frac{1}{4} \\ x^{2} + x + \frac{1}{4} & = - \frac{9}{4} \end{aligned}$
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
$\begin{aligned} \sqrt{{(x + \frac{1}{2})}^{2}} & = \pm \sqrt{- \frac{9}{4}} \\ x + \frac{1}{2} & = \pm \sqrt{- \frac{9}{4}} \end{aligned}$
Step 6. Move the constant term from left hand side to the right hand side, then solve for $x$
$\begin{aligned} x & = - \frac{1}{2} \pm \sqrt{- \frac{9}{4}} \\ x_{1} & = - \frac{1}{2} - \frac{3 i}{2} \\ x_{2} & = - \frac{1}{2} + \frac{3 i}{2} \end{aligned}$
This quadratic equation has 2 complex roots, as can be expected because the discriminant ( $b^{2} - 4 a c$ ) is negative.

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By Jimmy Sie

Solving Quadratic Equations by Completing the Square

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