- Completing The Square Definition
- Completing The Square
- How To Complete The Square
- Steps To Completing The Square
- Completing The Square Examples
- Solving Quadratic Equations

Algebra and geometry are closely connected. Geometry, as in coordinate graphing and polygons, can help you make sense of algebra, as in quadratic equations. **Completing the square** is one additional mathematical tool you can use for many challenges:

- Simplify algebraic expressions
- Solve quadratic equations
- Convert expressions from one form to another
- Find the minimum or maximum values of quadratic functions

When completing the square, we can take a quadratic equation like this, and turn it into this:

${a}{{x}}^{{2}}{+}{b}{x}{+}{c}{=}{0}\mathbf{\to}{a}{(}{x}{+}{d}{{)}}^{{2}}{+}{e}{=}{0}$

"**Completing the square**" comes from the exponent for one of the values, as in this simple **binomial expression**:

${x}^{2}+bx$

We use $b$ for the second term because we reserve $a$ for the first one. We might have had $a{x}^{2}$, but if $a$ is 1, you have no need to write it.

Anyway, you have no idea what values $x$ or $b$ have, so how can you proceed? You already know $x$ will be multiplied times itself, to begin.

Think about a square in geometry. You have four congruent-length sides, with an enclosed area that comes from multiplying a number times itself. In this expression, $x$ times $x$ is a square with an area of ${x}^{2}$:

Hold on -- we still have unknown variable $b$ times $x$. What would that look like? That would be a rectangle $x$ units tall and $b$ units wide, attached to our ${x}^{2}$ square:

To make better sense of that rectangle, divide it equally between the width and length of the ${x}^{2}$ square. That would make each rectangle $\frac{b}{2}$ times $x$:

That means the new almost-square is $x+\frac{b}{2}$, but we are missing a tiny corner, which would have a value of $\frac{b}{2}$ times itself, or ${\left(\frac{b}{2}\right)}^{2}$:

That last step literally completed the square, so now we have this:

${x}^{2}+bx+(\frac{b}{2}{)}^{2}$

**This refines or simplifies to:**

${\left(x+\frac{b}{2}\right)}^{2}$

You need to also subtract ${\left(\frac{b}{2}\right)}^{2}$ if you are, in fact, trying to work an equation (you cannot add something without balancing it by subtracting it). In our case, we were just showing how the square is really a square, in a geometric sense.

Here is a more complete version of the same thing:

${x}^{2}+2x+3$

As soon as you see $x$ raised to a power, you know you are dealing with a candidate for "completing the square."

The role of $b$ from our earlier example is played here by the $2$. We added a value, $+3$, so now we have a **trinomial expression**.

**${x}^{2}+2x+3$ is rewritten as:**

${x}^{2}+2bx+{b}^{2}$

**So, divide $b$ by $2$ and square it, which you then add and subtract to get:**

${x}^{2}+2x+3+{\left(\frac{2}{2}\right)}^{2}-{\left(\frac{2}{2}\right)}^{2}$

**Now, you can simplify as:**

${x}^{2}+2x+3+{1}^{2}-{1}^{2}$

**Which is equal to:**

${\left(x+1\right)}^{2}+3-{1}^{2}$

**This simplifies to:**

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

On a graph, this plots a parabola with a vertex at $\left(-1,2\right)$.

You can use completing the square to **simplify algebraic expressions**. Here is a straightforward example with steps:

${x}^{2}+20x-10$

Divide the middle term, $20x$, by $2$ and square it, then both add and subtract it:

${x}^{2}+20x-10+{\left(\frac{20}{2}\right)}^{2}-{\left(\frac{20}{2}\right)}^{2}$

**Simplify the expression:**

${x}^{2}+20x-10+{10}^{2}-{10}^{2}$

${\left(x+10\right)}^{2}-10-{10}^{2}$

${\left(\mathbf{x}\mathbf{+}\mathbf{10}\right)}^{\mathbf{2}}\mathbf{-}\mathbf{110}$

Seven steps are all you need to complete the square in any **quadratic equation**. The general form of a quadratic equation looks like this:

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

**Completing The Square Steps**

- Isolate the number or variable
**c**to the right side of the equation. - Divide all terms by
**a**(the coefficient of*x*^{2}, unless*x*^{2}has no coefficient). - Divide coefficient
**b**by two and then square it. - Add this value to both sides of the equation.
- Rewrite the left side of the equation in the form
**(x + d)**where^{2}**d**is the value of**(b/2)**you found earlier. - Take the square root of both sides of the equation; on the left side, this leaves you with
**x + d**. - Subtract whatever number remains on the left side of the equation to yield
**x**and**complete the square**.

We will provide three examples of quadratic equations progressing from easier to harder. Give each a try, following the seven steps described above. The first one does not place a coefficient with ${x}^{2}$:

- ${x}^{2}+3x-4=0$
- ${x}^{2}+3x=4$
- ${x}^{2}+3x+{\left(\frac{3}{2}\right)}^{2}=4+{\left(\frac{3}{2}\right)}^{2}$
- ${\left(x+\frac{3}{2}\right)}^{2}=\frac{25}{4}$
- $x+\frac{3}{2}=-\sqrt{\frac{25}{4}}$
- $x+\frac{3}{2}=\sqrt{\frac{25}{4}}$

$\mathbf{x}\mathbf{=}\mathbf{1}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{4}$

Our second example uses a coefficient with ${x}^{2}$ for solving a quadratic equation by completing the square:

- $2{x}^{2}-4x-2=0$
- $2{x}^{2}-4x=2$
- ${x}^{2}-2x=1$
- ${x}^{2}-2x+{\left(\frac{-2}{2}\right)}^{2}=1+{\left(\frac{-2}{2}\right)}^{2}$
- ${x}^{2}-2x+{\left(-1\right)}^{2}=1+{\left(-1\right)}^{2}$
- ${x}^{2}-2x+{\left(-1\right)}^{2}=2$
- ${\left(x-1\right)}^{2}=2$
- $x-1=-\sqrt{2}$
- $x-1=\sqrt{2}$

$\mathbf{x}\mathbf{=}\mathbf{-}\sqrt{\mathbf{2}\mathbf{+}\mathbf{1}}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{=}\sqrt{\mathbf{2}\mathbf{+}\mathbf{1}}$

Our third example is all bells and whistles with really big numbers. See how you do!

- $20{x}^{2}-30x-40=0$
- $20{x}^{2}-30x=40$
- ${x}^{2}-1.5x=2$
- ${x}^{2}-1.5x+{\left(\frac{-1.5}{2}\right)}^{2}=2+{\left(\frac{-1.5}{2}\right)}^{2}$
- ${x}^{2}-1.5x+{\left(0.75\right)}^{2}=2+{\left(0.75\right)}^{2}$
- ${x}^{2}-1.5x+{\left(-0.75\right)}^{2}=\frac{41}{16}$
- $(x-0.75{)}^{2}=\frac{41}{16}$
- $x-0.75=-\sqrt{\frac{41}{16}}$
- $x-0.75=\sqrt{\frac{41}{16}}$

$\mathbf{x}\mathbf{=}\frac{\left(\mathbf{-}\sqrt{\mathbf{41}\mathbf{+}\mathbf{3}}\right)}{\mathbf{4}}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{=}\frac{\left(\sqrt{\mathbf{41}\mathbf{+}\mathbf{3}}\right)}{\mathbf{4}}$

Instructor: **Malcolm M.**

Malcolm has a Master's Degree in education and holds four teaching certificates. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher.

Malcolm has a Master's Degree in education and holds four teaching certificates. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher.

Get better grades with tutoring from top-rated private tutors. Local and online.

View Tutors

Tutors online

Ashburn, VA

Get better grades with tutoring from top-rated professional tutors. 1-to-1 tailored lessons, flexible scheduling. Get help fast. Want to see the math tutors near you?

Learn faster with a math tutor. Find a tutor locally or online.

Get Started