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# Completing the Square

A polynomial equation with a degree equal to two is known as a quadratic equation. While "quad" means four, "quadratic" means to make a square. A quadratic equation in standard form is represented as:

${ax}^{2}+bx+c=0$

where b and c are real numbers such that a does not equal 0.

Since the degree of the above-written equation is two, it will have two roots, or solutions. The roots of polynomials are the values of x that satisfy the equation. There are a handful of methods you can use to find the roots of a quadratic equation.

One of them is called completing the square. Using this method, you have to convert the given equation into a perfect square.

## Steps to completing the square

To solve ${ax}^{2}+bx+c=0$ by completing the square:

1. Transform the equation so the constant term, c, is alone on the right side.
2. If a, the leading coefficient (the coefficient of the x2 term) is not equal to 1, divide both sides by a.
3. Add the square of half the coefficient of the x-term, (b/(2a))2 to both sides of the equation.
4. Factor the left side as the square of a binomial.
5. Take the square root of both sides. Remember that ${\left(x+q\right)}^{2}=r$ is equivalent to $x+q=±\sqrt{r}$ .
6. Solve for x.

The above steps can be implemented as shown below.

${ax}^{2}+bx+c=0$ (a not equal to 0)

Get c by itself on the right-hand side

${ax}^{2}+bx=-c$

Dividing both sides by a, we get

${x}^{2}+\frac{b}{a}x=-\frac{c}{a}$

Add ${\left(\frac{b}{2a}\right)}^{2}$ to both sides

${x}^{2}+\left(\frac{b}{a}\right)x+{\left(\frac{b}{2a}\right)}^{2}=-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}$

Factor the left-hand side

${\left(\frac{2ax+b}{2a}\right)}^{2}=-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}$

Square root both sides

$\frac{2ax+b}{2a}=\sqrt{-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}}$

This might look overwhelming, but recall that a, b, and c are just numbers, so the whole right side is just some constant.

Example 1

Solve ${x}^{2}-6x-3=0$ by completing the square.

${x}^{2}-6x=3$

${x}^{2}-6x+\left(-3\right){}^{2}=3+9$

${\left(x-3\right)}^{2}=12$

$x-3=±\sqrt{12}$

$x-3=±2\sqrt{3}$

$x=3±2\sqrt{3}$

Example 2

Solve $7{x}^{2}-8x+3=0$ by completing the square.

$7{x}^{2}-8x=-3$

${x}^{2}-\frac{8x}{7}=-\frac{3}{7}$

${x}^{2}-\frac{8x}{7}+{\left(-\frac{4}{7}\right)}^{2}=-\frac{3}{7}+\frac{16}{49}$

${\left(x-\frac{4}{7}\right)}^{2}=\frac{-5}{49}$

$x-\frac{4}{7}=\frac{\sqrt{5}}{7}i$

$x=\frac{4}{7}±\sqrt{\frac{\sqrt{5}}{7}i}$

## Alternate ways of writing the above steps

The above steps can also be written by defining 2 more terms d and e.

$d=\frac{b}{2a}$

$e=c-\frac{{b}^{2}}{4a}$

We can then write:

$a\left(x+d{\right)}^{2}+e=0$

and solve for x in the following way

${\left(x+d\right)}^{2}=-\frac{e}{a}$

$x+d=±\sqrt{\frac{-e}{a}}$

$x=-d±\sqrt{\frac{-e}{a}}$

Then we can either put in our values for a,b, and c, or d and e to get the answers for x.

$a\left(x+\frac{b}{2a}{\right)}^{2}+c-\frac{{b}^{2}}{4a}=0$

Further simplification of this will give you the quadratic formula, which is another method for solving quadratic equations.

## Practice questions on completing the square

1. Find the roots of the quadratic equation ${x}^{2}+4x-5=0$ by the method of completing the square.

${x}^{2}+4x-5=0$

Comparing the equation with the standard form,

$b=4,c=-5$

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

So, ${\left(x+\frac{4}{2}\right)}^{2}=-\left(5-\frac{{4}^{2}}{4}\right)$

${\left(x+2\right)}^{2}=5+4$

${\left(x+2\right)}^{2}=9$

$\left(x+2\right)=±\sqrt{9}$

$\left(x+2\right)=±3$

$x+2=3,x+2=-3$

$x=1,-5$

2. Find the roots of the quadratic equation $3{x}^{2}-5x+2=0$ by completing the square.

$3{x}^{2}-5x+2=0$

The given equation is not in the form to which we apply the method of completing the square because the coefficient of ${x}^{2}$ is not 1. To make it 1, we need to divide the whole equation by 3:

${x}^{2}-\frac{5}{3}x+\frac{2}{3}=0$

Comparing with the standard form,

$b=\frac{5}{3},c=\frac{2}{3}$

$c-\frac{{b}^{2}}{4}=\frac{2}{3}-\frac{{\left(-\frac{5}{3}\right)}^{2}}{4}=\frac{2}{3}-\frac{25}{36}=\frac{24-25}{36}=-\frac{1}{36}$

Substituting these values in the equation ${\left(x+\frac{b}{2}\right)}^{2}=-\left(c-\frac{{b}^{2}}{4}\right)$ , we get

${\left(x-\frac{5}{6}\right)}^{2}=\frac{1}{36}$

$\left(x-\frac{5}{6}\right)=±\sqrt{\frac{1}{36}}$

$\left(x-\frac{5}{6}\right)=±\frac{1}{6}$

$x=1,\frac{2}{3}$

## Flashcards covering the Completing the Square

Algebra 1 Flashcards

## Get help learning about completing the square

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