6.1 Solving Quadratic Equations

Learning Outcomes

Solve quadratic equations using the square root principle for equations with integer, rational, irrational, or complex number solutions.
Solve quadratic equations by the method of completing the square for equations with integer, rational, irrational, or complex number solutions.

Quadratic equations can be solved using many methods. You should already be familiar with factoring to solve some quadratic equations. However, not all quadratic equations can be factored. In this section, you will learn two other ways to solve quadratic equations.

Solve a Quadratic Equation by the Square Root Property

In this lesson we will solve quadratic equations where the variable is raised to the second power, $x^2$ . Examples of the types of quadratic equations that will be solved include:

$x^{2}=25$
$(x-3)^{2}=64$
$3(x+1)^{2}=48$

Whenever there is a way to do something in math, there is always a way to undo it. Every operation in mathematics has an inverse operation. The inverse operation of subtraction is addition. The inverse operation of multiplication is division. The inverse operation of taking the square root is squaring (raising to the power of 2). When we take the square root, we must remember to take both the positive AND the negative square roots.

One way to solve the quadratic equation $x^{2}=9$ is to subtract $9$ from both sides to get right side of the equation equal to 0.

$x^{2}-9=0$

Notice that the expression on the left can be factored; it is a difference of squares. Now you can solve this quadratic equation is by factoring the left side of the equation and then using the zero-product property.

$\left(x+3\right)\left(x–3\right)=0$ .

Using the zero-product property, $x+3=0$ or $x –3=0$ , so $x=−3$ or $3$ .

Another property that would let you solve this equation more easily is called the square root property.

The Square Root Property

If $x^{2}=p$ ,

then $x=\sqrt{p}$ or $x=-\sqrt{p}$ .

The property above says that you can take the square root of both sides of an equation, but you have to think about two cases: the positive square root of $p$ and the negative square root of $p$ .

When using the Square Root Property to solve an equation such as the one above ( $x^2=p$ ), the solutions can be abbreviated as $x=\pm \sqrt{p}$ . This is read as “x equals plus or minus the square root of p.” These still represent the same solutions of $x=\sqrt{p}$ or $x=-\sqrt{p}$ .

Example

Solve using the Square Root Property. $x^{2}=9$

Show Solution

Since one side of the equation already has $x^{2}$ isolated, you can take the square root of both sides to solve for $x$ .

$\begin{array}{c}x^{2}=9 \\ x=\pm\sqrt{9} \end{array}$

$x=\pm3$ (that is, $x=3$ or $-3$ )

Check your solutions:

For $x=3$ :

$\begin{align}\require{color}(\color{Green}{3}\color{black}{)^2-9} &=0\\ 9-9 &= 0\\ 0 &=0 && \color{Green}{TRUE}\end{align}$

For $x=-3$ :

$\begin{align}\require{color}(\color{Green}{-3}\color{black}{)^2-9} &=0\\ 9-9 &= 0\\ 0 &=0 && \color{Green}{TRUE}\end{align}$

They both solutions check. The solutions are $x=3$ or $x=-3$ .

Notice that there is a difference here in solving $x^{2}=9$ and simplifying $\sqrt{9}$ . For $x^{2}=9$ , you are looking for all numbers whose square is $9$ . For $\sqrt{9}$ , you only want the principal (nonnegative) square root.

In the example above, you can take the square root of both sides easily because there is only one term on each side. In some equations, you may need to do some work to get the equation in this form. You will find that this involves isolating $x^{2}$ .

In our first video, we will show more examples of using the square root property to solve a quadratic equation.

Example

Solve. $10x^{2}+5=85$

Show Solution

We need to isolate $x^2$ . Let’s do this by first subtracting $5$ from both sides.

$10x^{2}+5=85$

$\require{color}\hspace{8mm}\color{Green}{\underline{-5}\hspace{5mm}\underline{-5}}$

Next, let’s divide by $10$ to isolate $x^2$ .

$\dfrac{10x^{2}}{\color{Green}{10}}=\dfrac{80}{\color{Green}{10}}$

$x^{2}=8$

Now you have only $x^{2}$ on the left, so you can use the Square Root Property easily.

Be sure to simplify the radical if possible.

$\begin{align}{x}^{2} &= 8\\ x &= \pm \sqrt{8}\\ x &= \pm \sqrt{(4)(2)}\\ x &= \pm \sqrt{4}\sqrt{2}\\ x &= \pm 2\sqrt{2}\end{align}$

In the next example, we have a quantity squared and that is what we need to isolate to solve this using the Square Root Property.

Example

Solve. $\left(x–2\right)^{2}–50=0$

Show Solution

Start by adding 50 to both sides.

$\begin{align}\left(x–2\right)^{2}–50 &= 0\\ \left(x-2\right)^{2} &= 50\end{align}$

Because $\left(x–2\right)^{2}$ is a squared quantity, you can take the square root of both sides.

$\begin{align}{\left(x-2\right)}^{2} &= 50\\ \sqrt{{\left(x-2\right)}^{2}} &= \sqrt{50}\\ x-2 &= \pm\sqrt{50}\end{align}$

To isolate $x$ on the left, you need to add $2$ to both sides.

Be sure to simplify the radical.

$\begin{align}x &= 2\pm \sqrt{50}\\ x &= 2\pm \sqrt{(25)(2)}\\ x &= 2\pm \sqrt{25}\sqrt{2}\\ x &= 2\pm 5\sqrt{2}\end{align}$

The answer is $x=2\pm 5\sqrt{2}$ .

In the next video, you will see more examples of using square roots to solve quadratic equations.

Solve a Quadratic Equation by Completing the Square

When you square a number, the product is called a perfect square. For example, $5^2=5\cdot 5=25$ so 25 is a perfect square. Here is a list of some integers that are also perfect squares: $1,4,9,16,25,36,49,64,81,100...$ . We can also visualize integers that are perfect squares by arranging equal rows and columns to form a square, as shown below.

Take a closer look at the perfect square trinomials below. What do you notice about the constant terms? How are those constant terms related to the factored form and the binomial squared?

Perfect Square Trinomial	Factored Form	Binomial Squared
$\large{x^2+10x+\color{BurntOrange}{\fbox{25}}}$	$\large{(x+\color{BurntOrange}{5}\color{black}{)(x+}\color{BurntOrange}{5}\color{black}{)}}$	$\large{(x+\color{BurntOrange}{5}\color{black}{)^2}}$
$\large{x^2-6x+\color{BurntOrange}{\fbox{9}}}$	$\large{(x-\color{BurntOrange}{3}\color{black}{)(x-}\color{BurntOrange}{3}\color{black}{)}}$	$\large{(x-\color{BurntOrange}{3}\color{black}{)^2}}$
$\large{x^2-8x+\color{BurntOrange}{\fbox{16}}}$	$\large{(x-\color{BurntOrange}{4}\color{black}{)(x-}\color{BurntOrange}{4}\color{black}{)}}$	$\large{(x-\color{BurntOrange}{4}\color{black}{)^2}}$

Did you notice in the first example that the constant term, $25$ , of the trinomial is a perfect square? We know that $25=5\cdot 5=5^2$ and this helps us to confirm why there are $5$ ‘s in the factored form and the $5$ in the binomial squared form. Did you notice that the constant term of the trinomial is always positive, even when the $x$ term is negative? Notice that this same idea applies to the other two examples above.

Let’s take a look at the $x$ term (middle term) of the trinomial. How is it related to the binomial that is being squared? Looking at the first example above, did you notice that if you find half of the coefficient of the middle term of the trinomial, $10$ , that it is equal to $5$ ? Which is also the second term of the binomial squared. Does this same pattern hold true for the other above examples? Yes it does. In the second example, if we take half of $-6$ , we get $-3$ and in the third example if we take half of $-8$ we get $-4$ .

Not all trinomials can be factored and not all trinomials in the form of $ax^2+bx+c$ are perfect square trinomials. When they are not, we will need to change these trinomials into perfect square trinomials so that we can use this method to solve quadratic equations. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. In this course, we will only be completing the square on trinomials that have a leading coefficient of $1$ . In later courses you will learn how to complete the square when the leading coefficient does not equal 1.

First, let’s make sure we remember how to factor a perfect square trinomial.

Example

Factor $9x^{2}–24x+16$ .

Show Solution

First notice that the $x^{2}$ term and the constant term are both perfect squares. Also notice that the middle term (ignoring the sign) is twice the product of the square roots of these squared terms.

Below we have rewritten each term to make it easier to factor.

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

$24x=2\left(3x\right)\left(4\right)$

$16=4^{2}$

A trinomial in the form $m^{2}-2mn+n^{2}$ can be factored as $(m-n)^{2}$ .

In this case, the middle term is subtracted, so subtract $m$ and $n$ and square it to get $(m–n)^{2}$ .

$\begin{array}{c}\,\,\,m=3x\\n=4\\9x^{2}-24x+16=\left(3x-4\right)^{2}\end{array}$

If this were an equation, we could solve using either the square root property or the zero product property. If you do not start with a perfect square trinomial, you can complete the square to make what you have into one.

In this course when we complete the square, the leading coefficient, a, will equal 1. Let’s look at an equation that does not have a perfect square trinomial.

$x^2+8x-84=0$

To change the left side of the equation into a perfect square trinomial, let’s start by adding 84 to both sides of the equation which is:

$x^2+8x=84$ and rewrite the equation to look like $x^2+8x+\color{BurntOrange}{\_\_\_}\color{black}{= 84+}\color{BurntOrange}{\_\_\_}$ .

Now we need to figure out what to put in the blanks to make the left side of the equation a perfect square trinomial. To do this, we will find half of the coefficient of the middle term which is $\dfrac{8}{2}=4$ . Then take that answer and square it $(4)^2=16$ . So, $16$ is the number that we need to fill in the blanks with to make the left side into a perfect square trinomial.

$\begin{align}x^2+8x+\color{BurntOrange}{16} &= 84+\color{BurntOrange}{16}\\ x^2+8x+16 &= 100\end{align}$

Now, rewrite the left side of the equation as a binomial squared. Then square root both sides of the equation (using the Square Root Property) to solve for $x$ .

$\begin{align}(x+4)^2 &= 100\\ \sqrt{(x+4)^2} &= \sqrt{100}\\ x+4 &= \pm 10\end{align}$

Now write as two equations and solve each equation by subtracting $4$ from both sides of each of the equations.

$x+4=10\hspace{5mm}$ OR $\hspace{1cm}x+4=-10$

$x=6\hspace{1.5cm}$ OR $\hspace{1cm}x=-14$

You might be wondering why this method is called Completing the Square. The following video will help you visualize geometrically what is meant by Completing the Square.

We can use the following steps to solve a quadratic equation by completing the square.

Steps for Completing The Square

We will use the example ${x}^{2}+4x+1=0$ to illustrate each step.

Given a quadratic equation that cannot be factored and with $a=1$ , first add or subtract the constant term to the right side of the equal sign.
${x}^{2}+4x=-1$
Multiply the $b$ term by $\frac{1}{2}$ and square it.
$\begin{array}{c}\frac{1}{2}\left(4\right)=2\hfill \\ {2}^{2}=4\hfill \end{array}$
Add ${\left(\frac{1}{2}b\right)}^{2}$ to both sides of the equal sign and simplify the right side. In this case we will add 4, then we have:
$\begin{array}{l}{x}^{2}+4x+4=-1+4\hfill \\ {x}^{2}+4x+4=3\hfill \end{array}$
The left side of the equation can now be factored as a perfect square trinomial.
$\begin{array}{c}{x}^{2}+4x+4=3\hfill \\ {\left(x+2\right)}^{2}=3\hfill \end{array}$
Use the square root property and solve.
$\begin{array}{c}\sqrt{{\left(x+2\right)}^{2}}= \sqrt{3}\hfill \\ x+2=\pm \sqrt{3}\hfill \\ x=-2\pm \sqrt{3}\hfill \end{array}$
The exact solutions are $x=-2+\sqrt{3}$ , $x=-2-\sqrt{3}$ . Sometimes you might be asked to give approximated solutions to a given place value. The approximated solutions rounded to the nearest hundredth are $x\approx 3.73, x\approx 0.27$ .

Example

Solve by completing the square. $x^{2}–12x–4=0$

Show Solution

Since you cannot factor the trinomial on the left side, you will use completing the square to solve the equation.

First, move the constant term to the right side of the equal sign.

$\begin{array}{r}x^{2}-12x=4\end{array}$

What is $b$ ? $\hspace{5mm}b=-12$

Now take $\frac{1}{2}$ of the $b$ term and square it. Add ${{\left( \frac{b}{2}\right)}^{2}}$ to complete the square, so ${{\left( \frac{b}{2} \right)}^{2}}={{\left( \frac{-12}{2} \right)}^{2}}={{\left( -6 \right)}^{2}}=36$ .

Add the value to both sides of the equation and simplify.

$\begin{array}{l}x^{2}-12x+36=4+36\\x^{2}-12x+36=40\end{array}$

Rewrite the left side as a squared binomial.

$\left(x-6\right)^{2}=40$

Use the Square Root Property. Remember to include both the positive and negative square root, or you will miss one of the solutions.

$x-6=\pm\sqrt{40}$

Solve for $x$ by adding $6$ to both sides. Simplify as needed.

$\begin{array}{l}x & =6\pm \sqrt{40}\\ & =6\pm \sqrt{4}\sqrt{10}\\ & =6\pm 2\sqrt{10}\end{array}$

The exact solutions are $x=6\pm 2\sqrt{10}$ .

On this example, we have two irrational solutions.

Example

Solve by completing the square: ${x}^{2}-3x - 5=0$ .

Show Solution

First, move the constant term to the right side of the equal sign.

x^{2} - 3 x = 5

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

What is $b$ ? $b=-3$
Then, take $\frac{1}{2}$ of the $b$ term and square it.

\begin{matrix} \frac{1}{2} (- 3) = - \frac{3}{2} {(- \frac{3}{2})}^{2} = \frac{9}{4} \end{matrix}

$\begin{array}{l}\frac{1}{2}\left(-3\right)=-\frac{3}{2}\hfill \\ {\left(-\frac{3}{2}\right)}^{2}=\frac{9}{4}\hfill \end{array}$

Add the result to both sides of the equal sign.

\begin{matrix} x^{2} - 3 x + {(- \frac{3}{2})}^{2} = 5 + {(- \frac{3}{2})}^{2} x^{2} - 3 x + \frac{9}{4} = 5 + \frac{9}{4} \end{matrix}

$\begin{array}{l}\text{ }{x}^{2}-3x+{\left(-\frac{3}{2}\right)}^{2}=5+{\left(-\frac{3}{2}\right)}^{2}\hfill \\ {x}^{2}-3x+\frac{9}{4}=5+\frac{9}{4}\hfill \end{array}$

Factor the left side as a perfect square and simplify the right side.

{(x - \frac{3}{2})}^{2} = \frac{29}{4}

${\left(x-\frac{3}{2}\right)}^{2}=\frac{29}{4}$

Use the square root property and solve.

\begin{matrix} \sqrt{{(x - \frac{3}{2})}^{2}} & = \pm \sqrt{\frac{29}{4}} x - \frac{3}{2} & = \pm \frac{\sqrt{29}}{2} x & = \frac{3}{2} \pm \frac{\sqrt{29}}{2} \end{matrix}

$\begin{array}{rl}\sqrt{{\left(x-\frac{3}{2}\right)}^{2}}\hfill & = \pm \sqrt{\frac{29}{4}}\hfill \\ x-\frac{3}{2} & =\pm \frac{\sqrt{29}}{2}\hfill \\ x & =\frac{3}{2}\pm \frac{\sqrt{29}}{2}\hfill \end{array}$

The solutions are $x=\frac{3+\sqrt{29}}{2}$ , $x=\frac{3-\sqrt{29}}{2}$ .

We end up with two irrational solutions.

In the next video, you will see more examples of how to use completing the square to solve a quadratic equation.

You may have noticed that because you have to use both square roots, all the examples have two solutions. Here is another example that is slightly different.

Example

Solve by completing the square. $x^{2}+16x+17=-47$ .

Show Solution

Rewrite the equation so the left side has the form $x^{2}+bx$ . Identify b.

$\begin{array}{c}x^{2}+16x=-64\\b=16\end{array}$

Add ${{\left( \frac{b}{2} \right)}^{2}}$ , which is ${{\left( \frac{16}{2} \right)}^{2}}={{8}^{2}}=64$ , to both sides.

$\begin{array}{l}x^{2}+16x+64=-64+64\\x^{2}+16x+64=0\end{array}$

Write the left side as a squared binomial.

$\left(x+8\right)^{2}=0$

Take the square roots of both sides. Normally both positive and negative square roots are needed, but 0 is neither positive nor negative. $0$ has only one real solution.

$x+8=0$

$x=-8$

Take a closer look at this problem and you may see something familiar. Instead of completing the square, try adding $47$ to both sides in the equation. The equation $x^{2}+16x+17=−47$ becomes $x^{2}+16x+64=0$ . Can you factor this equation using grouping? (Think of two numbers whose product is 64 and whose sum is $16$ ).

It can be factored as $(x+8)(x+8)=0$ , of course! Knowing how to complete the square is very helpful, but it is not always the only way to solve an equation.

In our last video, we show an example of how to use completing the square to solve a quadratic equation whose solutions are irrational.

In the next example, take notice of the type and number of solutions.

Example

Solve by completing the square. $x^{2}-2x-41=-60$ .

Show Solution

First, move the constant term to the right side of the equal sign.

x^{2} - 2 x = - 19

${x}^{2}-2x=-19$

What is $b$ ? $b=-2$
Then, take $\frac{1}{2}$ of the $b$ term and square it.

$\begin{array}{l}\frac{1}{2}\left(-2\right)=\frac{-2}{2}=-1\hfill \\ (-1)^{2}=1\hfill \end{array}$

Add the result to both sides of the equal sign.

$\begin{array}{l}x^{2}-2x+1=-19+1\\x^{2}-2x+1=-18\end{array}$

Rewrite the left side as a squared binomial. Then use the Square Root Property to solve.

$\left(x-1\right)^{2}=-18$

$\sqrt{(x-1)^{2}}=\sqrt{-18}$

$x-1=\pm\sqrt{-18}$

$x-1=\pm 3i \sqrt{2}$

Solve for $x$ by adding $1$ to both sides. Simplify as needed.

$x=1\pm 3i\sqrt{2}$

The solutions can also be written as $x=1\pm 3\sqrt{2}i$ . If you write it this way, make sure the $i$ is not under the square root.
The exact solutions are $x=1+ 3i\sqrt{2}, x=1-3i\sqrt{2}$ .

We end up with two complex solutions.

Summary

Completing the square is used to change a binomial of the form $x^{2}+bx$ into a perfect square trinomial ${{x}^{2}}+bx+{{\left( \frac{b}{2} \right)}^{2}}$ which can be factored to ${{\left( x+\frac{b}{2} \right)}^{2}}$ . When we have a quadratic equation that cannot be factored, we can use completing the square to solve it. When solving quadratic equations by completing the square, be careful to add ${{\left( \frac{b}{2} \right)}^{2}}$ to both sides of the equation to maintain equality. The Square Root Property can then be used to solve for $x$ . With the Square Root Property, be careful to include both the principal square root and its opposite. Be sure to simplify as needed.

Module 6: Quadratic Functions and Equations

Learning Outcomes

Solve a Quadratic Equation by the Square Root Property

The Square Root Property

Example

Example

Example

Solve a Quadratic Equation by Completing the Square

Perfect Square Trinomial

Factored Form

Binomial Squared

Example

Steps for Completing The Square

Example

Example

Example

Example

Summary

Candela Citations