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Quadratic equations

Quadratic equation

Here you will learn about quadratic equations and how to solve quadratic equations using four methods: factoring, using the quadratic formula, completing the square and using a graph.

Students will first learn about quadratic equations as part of geometry in high school.

What is a quadratic equation?

A quadratic equation is a quadratic expression that is equal to something. Quadratic equations are a type of polynomial equation because they consist of algebraic terms, with the highest being second-degree.

A quadratic equation can have zero, one or two solutions. To solve a quadratic equation it must equal $0.$

For example,

\begin{aligned}& x^2=0 \\\\ & x^2-2 x=0 \\\\ & 2 x^2+3 x-2=0\end{aligned}

The standard form of the quadratic equation is:

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

$a$ is the coefficient (number in front) of the $x^2$ term.

$b$ is the coefficient (number in front) of the $x$ term.

$c$ is the constant term (number on its own).

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How to solve quadratic equations

In order to solve a quadratic equation, you must first check that it is in the form

$a x^{2}+b x+c=0.$ If it isn’t, you will need to rearrange the equation.

Example:

Let’s explore each of the four methods of solving quadratic equations by using the same example: $x^{2}-2x-24=0$

Step-by-step guide: Solving quadratic equation

a) Quadratic graphs

Solve $x^{2}-2x-24=0$ by using a quadratic graph.

The real roots/solutions are shown where the graph crosses the horizontal $x$ -axis.

$x=6 \;$ and $\; x=-4$

Step-by-step guide: Solving quadratic equations graphically

b) Completing the square

Solve $x^{2}-2x-24=0$ by completing the square.

Most quadratic expressions, including this one, are not perfect squares. When you complete the square, you try to fit the expression to the closest possible perfect square, by adding or subtracting to make things work.

First, move $c$ to the right side of the equation.

\begin{aligned} x^2-2 x-24&=0 \\\\ +24 & \;\; +24 \\\\ x^2-2 x&=24 \end{aligned}

Then, complete the square and add the square’s $c$ to both sides of the equation.

Since $(x-1)^2= x^2-2x +1,$ add $1$ to both sides.

\begin{aligned} x^2-2 x+1&=24+1 \\\\ x^2-2 x+1&=25 \\\\ (x-1)^2&=25\end{aligned}

Finally, solve for $x.$

(x-1)^2=25

$\sqrt{(x-1)^2}=\sqrt{25} \;$ *Take the square root of both sides

$x-1= \pm \, 5 \;$ *Solve for $-5$ and $5$

$\begin{aligned} x-1&=5 \\ +1 &\;\; +1 \\ x &=6 \quad \text{and} \end{aligned}$ $\begin{aligned} x +1 &=5 \\ -1 & \;\; -1 \\ x &=-4 \end{aligned}$

See also: Completing the square

c) Factoring

Solve $x^{2}-2x-24=0$ by factoring.

(x-6)(x+4)=0

\begin{aligned} x-6&=0 \hspace{1cm} x+4=0 \\\\ x&=6 \hspace{1.55cm} x=-4 \end{aligned}

Step-by-step guide: Factoring quadratic equations

d) Quadratic formula

You can substitute the values of $a, b$ and $c$ from the general form of the quadratic equation, $a x^2+b x+c=0,$ into the quadratic formula to calculate the solution(s) for the quadratic formula, $x.$

$x=\cfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Solve $x^{2}-2x-24=0$ using the quadratic formula.

a=1, \, b=-2, \, c=-24

Now, let’s solve for the solutions.

$\begin{aligned}& x=\cfrac{-(-2)+\sqrt{(-2)^2-4 \times 1 \times(-24)}}{2 \times 1} \\\\ & x=\cfrac{2+\sqrt{4-(-96)}}{2} \\\\ & x=\cfrac{2+10}{2} \\\\ & x=6\end{aligned} \hspace{0.3cm}$ $\hspace{0.3cm} \begin{aligned}& x=\cfrac{-(-2)-\sqrt{(-2)^2-4 \times 1 \times(-24)}}{2 \times 1} \\\\ & x=\cfrac{2-\sqrt{4-(-96)}}{2} \\\\ & x=\cfrac{2-10}{2} \\\\ & x=-4\end{aligned}$

Step-by-step guide: Quadratic formula

What is a quadratic equation?

Common Core State Standards

How does this relate to high school math?

Algebra – Reasoning with Equations and Inequalities (HSA.REI.B.4a)
Use the method of completing the square to transform any quadratic equation in $x$ into an equation of the form $(x-p)^2=q$ that has the same solutions. Derive the quadratic formula from this form.

Algebra – Reasoning with Equations and Inequalities (HSA.REI.B.4b)
Solve quadratic equations in one variable. Solve quadratic equations by inspection (e.g., for $x^2=49$ ), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation.

Recognize when the quadratic formula gives complex solutions and write them as $a \pm bi$ for real numbers $a$ and $b.$

How to solve quadratic equations

There are a lot of ways to solve quadratic equations. For more specific step-by-step guides, check out the quadratic equation pages linked in the “What is a quadratic equation?” section above or read through the examples below.

Quadratic equation examples

Example 1: solve using a graph

Solve $x^{2}-2 x-24=0$ using a graph.

Graph the quadratic equation.

2Identify where the graph crosses the $\textbf{x}$ -axis (when $\bf{\textbf{y}=0}$ )

You know that to solve a quadratic equation, it must be equal to $0.$ Since $x^2-2 x-24=0$ you are looking for the values of $x$ that when substituted into the equation will give us a $y$ value of $0.$

On the graph the coordinates for $x$ where $y=0$ are given where the graph crosses the $x$ axis.

x=6 \hspace{1cm} x=-4

So the solutions or roots of the equation $x^{2}-2 x-24=0$ are $x=6$ and $x=-4.$

You can check that the solution is correct by substituting it into the original equation.

Example 2: solve by completing the square

Solve $x^2-4x=12$ by completing the square.

Complete the square to rewrite the quadratic equation in the form.

a x^2+b x=c

The equation given is already in this form.

Complete the square and add the square’s $\textbf{c}$ to both sides of the equation.

Since $(x-2)^2=x^2-4 x+4,$ add $4$ to both sides.

$\begin{aligned} x^2-4 x+4&=12+4 \\\\ (x-2)^2&=16\end{aligned}$

Rearrange the equation to solve for $\textbf{x}.$

\begin{aligned} (x-2)^2&=16 \\\\ \sqrt{(x-2)^2}&=\sqrt{16} \\\\ x-2&= \pm \, 4\end{aligned}

The square root has a $\textbf{+}$ and $\textbf{−}$ answer. Write down both versions of the calculation to find the two solutions of $\textbf{x}$ .

$\begin{aligned}x-2&=4 \\ +2 &\;\; +2 \\ x&=6 \quad \text{and} \end{aligned}$ $\begin{aligned}x-2&=-4 \\ +2 & \;\; +2 \\ x&=-2 \end{aligned}$

You can check that our solution is correct by substituting it into the original equation.

Example 3: solve by completing the square

Solve $x^2+10x+18.75=0$ by completing the square.

Complete the square to rewrite the quadratic equation in the form.

a x^2+b x=c

Subtract $18.75$ from both sides.

$\begin{aligned} x^2+10 x+18.75&=0 \\\\ -18.75 &\;\; -18.75 \\\\ x^2+10 x&=-18.75 \end{aligned}$

Complete the square and add the square’s $\textbf{c}$ to both sides of the equation.

Since $(x+5)^2=x^2+10 x+25,$ add $25$ to both sides.

$\begin{aligned} x^2+10 x+25&=-18.75+25 \\\\ (x+5)^2&=6.25\end{aligned}$

Rearrange the equation to solve for $\textbf{x}.$

\begin{aligned} (x+5)^2&=6.25 \\\\ \sqrt{(x+5)^2}&=\sqrt{6.25} \\\\ x+5&= \pm \, 2.5\end{aligned}

The square root has a $\textbf{+}$ and $\textbf{−}$ answer. Write down both versions of the calculation to find the two solutions of $\textbf{x}.$

$\begin{aligned}x+5&=-2.5 \\ -5 &\;\; -5 \\ x&=-7.5 \quad \text{and} \end{aligned}$ $\begin{aligned}x+5&=2.5 \\ -5 &\;\; -5 \\ x&=-2.5 \end{aligned}$

You can check that our solution is correct by substituting it into the original equation.

Example 4: solve by factoring

Solve $x^{2}-2x-24=0$ by factoring.

Fully factor the quadratic equation.

You can factor a quadratic into two factors when it is in the form $a x^2+b x+c.$

When $a=1,$ find the factors of $c$ and see which to have the sum of $b.$

Set each factor equal to $\bf{0}.$

x-6=0 \hspace{1cm} x+4=0

Solve each equation to find $\textbf{x}.$

$\begin{aligned}x-6&=0 \\ +6 &\;\; +6 \\ x&=6\end{aligned} \quad$ $\quad \begin{aligned}x+4&=0 \\ -4 &\;\; -4 \\ x&=-4 \end{aligned}$

You can check that our solution is correct by substituting it into the original equation.

Example 5: solve by factoring

Solve $x^2+9x-10=0$ by factoring.

Fully factor the quadratic equation.

You can factor a quadratic into two factors when it is in the form $a x^2+bx+c.$

When $a=1,$ find the factors of $c$ and see which to have the sum of $b.$

Set each factor equal to $\bf{0}.$

x-1=0 \hspace{1cm} x+10=0

Solve each equation to find $\textbf{x}.$

$\begin{aligned}x-1&=0 \\ +1 &\;\; +1 \\ x&=1\end{aligned} \quad$ $\quad \begin{aligned}x+10&=0 \\ -10 & \;\; -10 \\ x &=-10 \end{aligned}$

You can check that our solution is correct by substituting it into the original equation.

Example 6: quadratic equation with complex roots

Solve $x^2+2x+2=0.$

Identify the value of $\textbf{a, b}$ and $\textbf{c}$ in a quadratic equation.

\begin{aligned}& a=1 \\\\ & b=2 \\\\ & c=2\end{aligned}

Substitute these values into the quadratic formula.

\begin{aligned}& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\ & x=\cfrac{-2 \pm \sqrt{2^2-4(1)(2)}}{2(1)}\end{aligned}

Solve the equation with a $\textbf{+}$ , and then with a $\textbf{−}$ .

$\begin{aligned}& x=\cfrac{-2+\sqrt{2^2-4(1)(2)}}{2(1)} \\\\ & x=\cfrac{-2+\sqrt{4-8}}{2} \\\\ & x=\cfrac{-2+2 i}{2} \\\\ & x=-1+i\end{aligned} \hspace{1cm}$ $\hspace{1cm} \begin{aligned}& x=\cfrac{-2-\sqrt{2^2-4(1)(2)}}{2(1)} \\\\ & x=\cfrac{-2-\sqrt{4-8}}{2} \\\\ & x=\cfrac{-2-2 i}{2} \\\\ & x=-1-i\end{aligned}$

Note, since $\sqrt{4-8}=\sqrt{-4},$ you use a complex number to represent the square root.

The roots are imaginary numbers, because this parabola does not cross the $x$ -axis.

Teaching tips for quadratic equation

Allow students use a graphing calculator to input different values for $a, b$ and $c$ in a quadratic equation and explore how changing the values changes the parabola.

After graphing, many curriculums start with the square method, as it is usually the easiest for students to understand when beginning to solve quadratic equations. However, always do what is best for your students.

Allow students who are struggling to use a calculator for all computations, since these are not the focus of this skill.

There are many free quadratic equation solvers on the internet. Students can use these to check their work when solving with any method.

Easy mistakes to make

Forgetting to calculate both solutions for the quadratic formula
The quadratic formula includes an operation with both a positive sign and negative sign. This is because it involves the calculation of the square root of the discriminant. Don’t forget to calculate both solutions.

Thinking that quadratic equations only look like $\bf{\textbf{ax}^2+\textbf{bx}+\textbf{c}=0}$
A quadratic equation is any equation where the highest term is $x^2$ (or any second-order polynomial). Quadratic equations do not have to come in the form $a x^2+b x+c=0,$ but since they are algebraic equations they can always be simplified to this form.
For example,
Simplify $3 x^2-3 x=2 x^2+14$ to the form $ax^2+bx+c=0.$
$3 x^2-3 x=2 x^2+14$
$\quad -2 x^2 \quad -2 x^2$
$\;\; x^2-3 x=14$
$\quad -14 \quad -14$
$x^2-3 x-14=0$

Forgetting to move all parts to the left-hand side before using the quadratic formula
The values for $a, b$ and $c$ in the quadratic formula will only work for a quadratic expression set equal to $0.$
For example,
Before using the quadratic equation to solve $4 x^2-9 x=12,$ you need to subtract $12$ from both sides so that the equation is in the form $4 x^2-9 x-12=0.$

Practice quadratic equation questions

x=-4, \; y=2

x=-1.5, \; x=1

x=-4, \; x=2

x=1, \; y=-1.5

Quadratic Equation 7 US

The roots of a quadratic equation are where the parabola passes through the $x$ -axis. In this graph, it passes at $(-1.5,0)$ and $(1,0).$

The solutions are $x = -1.5$ and $x = 1.$

x=0, \; x=-9

x=4, \; x=-4

x=4

x=9, \; x=-1

Complete the square and add the square’s $c$ to both sides of the equation.

Since $(x-4)^2=x^2-8 x+16,$ add $16$ to both sides.

\begin{aligned} x^2-8 x+16&=9+16 \\\\ (x-4)^2&=25\end{aligned}

Rearrange the equation to solve for $x.$

\begin{aligned} (x-4)^2&=25 \\\\ \sqrt{(x-4)^2}&=\sqrt{25} \\\\ x-4&= \pm \, 5\end{aligned}

$\begin{aligned}x-4&=5 \\ +4 & \;\; +4 \end{aligned} \quad$ $\quad \begin{aligned}x-4&=-5 \\ +4 & \;\; +4 \end{aligned}$

$\quad \;\; x=9 \quad$ and $\;\, \quad x=-1$

x=-0.05, \; x=-19.95

x=-10, \; x=-99

x=0

x=0.05, \; x=-0.05

First, solve so that $c$ is on the right-hand side of the equation.

\begin{aligned} x^2+20 x+1&=0 \\\\ -1 & \;\;-1 \\\\ x^2+20 x&=-1 \end{aligned}

Then, complete the square and add the square’s $c$ to both sides of the equation.

Since $(x+10)^2=x^2+20 x+100,$ add $100$ to both sides.

\begin{aligned} x^2+20 x+100&=-1+100 \\\\ (x+10)^2&=99\end{aligned}

Rearrange the equation to solve for $x.$

\begin{aligned} (x+10)^2&=99 \\\\ \sqrt{(x+10)^2}&=\sqrt{99} \end{aligned}

$\quad \quad x+10= \pm \, 9.95 \quad$ *rounded to the nearest hundredth

$\begin{aligned}x+10 &= 9.95 \\ -10 & \;\; -10 \\ x&=-0.05 \quad \; \text{and} \end{aligned}$ $\begin{aligned}x+10 &=-9.95 \\ -10 & \;\; -10 \\ x&=19.95\end{aligned}$

x=-9.1, \; x=1.9

x=-2, \; x=-18

x=1, \; x=-5

x=-1, \; x=-18

Find the factors of $ac,$ which is $2 \times-10=-20.$

\begin{aligned}& -1 \times 20=-20 \\\\ & -20 \times 1=-20 \\\\ & -2 \times 10=-20 \\\\ & -10 \times 2=-20 \\\\ & -4 \times 5=-20 \\\\ & -5 \times 4=-20\end{aligned}

The sum of $ac$ should be $b,$ or $8.$

$-2 +10=8,$ so use these to write an equivalent equation.

2 x^2+10 x-2 x-10

Then group the terms and factor out the $GCF.$

\begin{aligned}& \left(2 x^2+10 x\right)+(-2 x-10) \\\\ & 2 x(x+5)+[-2(x+5)] \\\\ & (2 x-2)(x+5)\end{aligned}

$\begin{aligned}2 x-2&=0 \\ +2 & \;\; +2 \\ 2x&=2 \\ x &=1 \quad \text{and} \end{aligned}$ $\begin{aligned}x+5&=0 \\ -5 & \;\; -5 \\ & \\ x&=-5\end{aligned}$

Note, that the factoring could also result in $(2 x+10)(x-1),$ which leads to the same roots.

x=2 \quad x=\cfrac{1}{3}=0.333…

x=-2 \quad x=\cfrac{1}{3}=0.333…

x=2 \quad x=-\cfrac{1}{3}=-0.333…

x=-2 \quad x=-\cfrac{1}{3}=-0.333…

Identify $a, b$ and $c.$

a=3, \; b=5, \; c=-2

Substitute these values into the quadratic formula.

\begin{aligned}& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\ & x=\cfrac{-5 \pm \sqrt{5^2-4(3)(-2)}}{2(3)}\end{aligned}

Solve the equation with a $+,$ and then with a $-.$

$\begin{aligned}& x=\cfrac{-5+\sqrt{5^2-4(3)(-2)}}{2(3)} \\\\ & x=\cfrac{-5+\sqrt{25-(-24)}}{6} \\\\ & x=\cfrac{-5+7}{6} \\\\ & x=\cfrac{1}{3}\end{aligned} \hspace{1cm}$ $\hspace{1cm} \begin{aligned}& x=\cfrac{-5-\sqrt{5^2-4(3)(-2)}}{2(3)} \\\\ & x=\cfrac{-5-\sqrt{25-(-24)}}{6} \\\\ & x=\cfrac{-5-7}{6} \\\\ & x=-2\end{aligned}$

Quadratic Equation 8 US

x=-0.25+(-1.2), \; x=-0.25-(-1.2)

x=-1.5 i, \; x=1 i

x=-1.5, \; x=1

x=-\cfrac{1}{4}-1.199 i, \; x=-\cfrac{1}{4}+1.199 i

Identify $a, b$ and $c.$

a=-4, \; b=2, \; c=-6

Substitute these values into the quadratic formula.

\begin{aligned}& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\ & x=\cfrac{2 \pm \sqrt{2^2-4(-4)(-6)}}{2(-4)}\end{aligned}

Solve the equation with a $+,$ and then with a $-.$

$\begin{aligned}& x=\cfrac{2+\sqrt{2^2-4(-4)(-6)}}{2(-4)} \\\\ & x=\cfrac{2+\sqrt{4-96}}{-8} \\\\ & x=\cfrac{2+\sqrt{92} i}{-8} \\\\ & x=-\cfrac{1}{4}-1.199 i\end{aligned} \hspace{1cm}$ $\hspace{1cm} \begin{aligned}& x=\cfrac{2-\sqrt{2^2-4(-4)(-6)}}{2(-4)} \\\\ & x=\cfrac{2-\sqrt{4-96}}{-8} \\\\ & x=\cfrac{2-\sqrt{92 i}}{-8} \\\\ & x=-\cfrac{1}{4}+1.199 i\end{aligned}$

Quadratic Equation 9 US

You can see that there are no values of $x$ that give a $y$ value of $0.$

The graph does not cross the $x$ -axis. The use of the quadratic formula will result in roots that are not real.

Quadratic equation FAQs

How can you find the number of solutions of the quadratic equation?

Calculate the discriminant $\left(b^2-4 a c\right).$ If the discriminant is equal to $0,$ it has $1$ distinct root. This is because the parabola’s vertex lies on the $x$ -axis, so it only crosses once.

If the discriminant is greater than $0,$ it has $2$ distinct roots that are real solutions. If the discriminant is less than $0,$ it has $2$ distinct roots that are complex solutions.

What are possible values for $\textbf{b}$ and $\textbf{c}$ in a quadratic function?

The values for $b$ and $c$ can be any real number, including rational numbers, fractions, decimals, integers or whole numbers.

The next lessons are

Radicals
Algebraic fractions
Vectors

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