# Completing the Square Practice Math Problems and Examples

Created by Anna Szczepanek, PhD
Reviewed by Rijk de Wet
Last updated: Jan 27, 2022

Here you can find practice questions for the method of solving quadratic equations by completing the square. This is one of the most important problems in high school math! Go through the examples and problems we offer, and become a master of completing the square.

## Solving general quadratic equations by completing the square

In high school algebra, you're bound to encounter the problem of solving general quadratic equations by the "completing the square" method. A general quadratic equation is an equation involving a quadratic polynomial (so a polynomial of degree two):

$ax^2 + bx + c = d$

where $a$, $b$, $c$ and $d$ are real coefficients. It's clear that such an equation can be simplified to get $a=1$ as the leading coefficient and to get the constant term at the right side of the equation equal to $0$. To do that, let's subtract $d$ from both sides and then divide by $a$:

$\begin{split} ax^2 + bx + c &= d \\ ax^2 + bx + (c-d) &= 0 \\ x^2 + \frac{b}{a} x + \frac{c-d}{a} &= 0 \\ \end{split}$

Consequently, it suffices to know how to solve a quadratic equation of the form:

$x^2+bx+c = 0$

To do it, add $-c+\frac{b^2}{4}$ to both sides:

$x^2+bx+\frac{b^2}{4} = -c+\frac{b^2}{4}$

We can see a perfect square trinomial on the left side:

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

Thus,

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

And so now, if the right side is non-negative, we take the square root of both sides — the left side of the equation becoming $|x + \frac{b}{2}|$ — and perform standard calculations to solve the equation. If the right side is negative, the equation has no real solution (although it does have complex solutions).

## Completing the square practice problems quiz

What term do you have to add to the following expression in order to get a perfect square trinomial?

1. $x^2 + 2x$
2. $x^2 - 6x$
3. $x^2 + 3x$
4. $x^2 + 6x +6$
5. $x^2 - 2x + 3$

1. Add $1$ to get the trinomial $(x+1)^2$.
2. Add $9$, trinomial: $(x-3)^2$.
3. Add $\frac{9}{4}$, trinomial: $(x+\frac{3}{2})^2$.
4. Add $3$, trinomial: $(x+3)^2$.
5. Subtract $2$, trinomial: $(x-1)^2$.

## Example questions

Here we solve quadratic equations by completing the square so that you can learn this method with some examples.

Example 1. $x^2 - x + 0.25 = 1$

We can expand the left-hand side as

$\begin{split} x^2 - x + 0.25 &= (x-0.5)^2 \\ \therefore (x-0.5)^2 &= 1 \\ \end{split}$

We can apply the square root to both sides to get $|x - 0.5|=1$ and therefore we know $x - 0.5 = \pm 1$. Therefore, $x = 1.5$ or $x = -0.5$.

Example 2. $2x^2 + 4x + 8 = 0$

Let's divide both sides by $2$:

$x^2 + 2x = 0$

Now, we examine the expression $x^2 + 2x$. To produce these terms by the short multiplication formula, we can use $(x + 1)^2 = x^2 + 2x + 1$. As you can see, the constant term is $1$ while in the equation it is $0$ — we need to complete the square by adding $1$:

$\begin{split} x^2 + 2x + 1 &= 1 \\ (x +1)^2 &= 1 \\ |x + 1| &= \sqrt{1} = 1 \\ x + 1 &= \pm 1 \\ \therefore \qquad x &= 0 \\ \text{or } x &= -2 \\ \end{split}$

Example 3. $x^2 - 8x + 20 = 0$

At the left side we have $x^2 - 8x$, which can be interpreted as a part of the perfect square trinomial $x^2 - 8x + 16 = (x-4)^2$. However, we have $20$ in our equation and only $16$ in the perfect square trinomial. So let's subtract $4$ from both sides:

$\begin{split} x^2 - 8x + 20 - 4 &= -4 \\ x^2 - 8x + 16 &= -4 \\ (x - 4)^2 &= -4 \\ \end{split}$

Clearly, this equation has no solution in real numbers, because no real number can give $-4$ when squared. Let's solve this equation in complex numbers (if you're not yet familiar with complex numbers, feel free to skip this part). So, we have

$\begin{split} (x - 4)^2 &= (2i)^2 \\ x-4 &= \pm 2i \\ x &= 4 \pm 2i \\ \end{split}$

## Practice questions

Solve using the method of completing the square:

1. $x^2 + 8x - 9 = 0$
2. $3x^2 + 12x = 0$
3. $x^2 + 4x = -4$

1. $x = -9$ or $x = 1$
2. $x = 0$ or $x = -4$
3. $x = -2$
Anna Szczepanek, PhD
We will solve the quadratic equation
ax2 + bx + c = 0
Enter your respective coefficients:
a
b
c
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