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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.

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).

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

**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}

**Step-by-step guide:** 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}**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=-24Now, 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

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.

Use this quiz to check your grade 6 – grade 8 students’ understanding of algebra. 10+ questions with answers covering a range of 6th to 8th grade algebra topics to identify areas of strength and support!

DOWNLOAD FREEUse this quiz to check your grade 6 – grade 8 students’ understanding of algebra. 10+ questions with answers covering a range of 6th to 8th grade algebra topics to identify areas of strength and support!

DOWNLOAD FREEThere 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.

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

**Graph the quadratic equation.**

2**Identify 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=-4So 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.

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.

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.

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.

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.

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.

- 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.

**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.

1. Solve 2 x^2+x-3 by using its graph below.

x=-4, \; y=2

x=-1.5, \; x=1

x=-4, \; x=2

x=1, \; y=-1.5

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.

2. Complete the square to solve the given quadratic equation:

x^2-8 x=9

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

3. Complete the square to solve the given quadratic equation:

x^2+20 x+1=0

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}

4. Solve by factoring 2 x^2+8 x-10=0.

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.

5. Solve 3{x}^2+5x-2=0 using the quadratic formula.

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}

6. Solve -4 x^2+2 x-6=0 using the quadratic formula.

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}

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.

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.

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

- Functions in algebra
- Laws of exponents
- Algebraic fractions

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