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Here you will learn about solving quadratic equations and how to do it using a graph, factoring the equation or using the quadratic formula.

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

**Solving quadratic equations** is finding the roots or the x- intercepts of the parabola formed by a quadratic equation.

For example,

Solve 2x^2+7x-4 by using its graph below.

The parabola passes through the x- axis as (-4,0) and \left(\cfrac{1}{2}, 0\right).

The solutions are x=-4 and x=\cfrac{1}{2}.

For example,

Solve 2x^2+7x-4=0 by factoring.

\begin{aligned} & 2 x^2+7 x-4=0 \\\\ & (2 x-1)(x+4)=0 \end{aligned}Find where each value being multiplied will be equal to 0.

2x-1=0, so x=\cfrac{1}{2}

and

x+4=0 so x=-4

For example,

Solve 2x^2+7x-4=0 using the quadratic formula.

The quadratic formula uses the values from the constant terms a, b and c when quadratic equations are in the general form ax^2+b x+c=0.

\begin{aligned} & a=2 \\\\ & b=7 \\\\ & c=-4 \end{aligned}\begin{aligned} & x=\cfrac{-b+\sqrt{b^2-4 a c}}{2 a} \\\\ & x=\cfrac{-7+\sqrt{7^2-4 \times 2 \times(-4)}}{2 \times 2} \\\\ & x=\cfrac{-7+\sqrt{49-(-32)}}{4} \\\\ & x=\cfrac{-7+9}{4} \\\\ & x=\cfrac{2}{4} \\\\ & x=\cfrac{1}{2} \end{aligned} \quad \begin{aligned} & x=\cfrac{-b-\sqrt{b^2-4 a c}}{2 a} \\\\ & x=\cfrac{-7-\sqrt{7^2-4 \times 2 \times(-4)}}{2 \times 2} \\\\ & x=\cfrac{-7-\sqrt{49-(-32)}}{4} \\\\ & x=\cfrac{-7-9}{4} \\\\ & x=\cfrac{-16}{4} \\\\ & x=-4 \end{aligned}

How does this relate to high school math?

**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 FREEIn order to solve quadratic equations by graphing:

**Graph the quadratic equation.****Identify the \textbf{x}- intercepts.**

Solve the following equation by using a graph: x^2-x-6.

**Graph the quadratic equation.**

2**Identify the \textbf{x}- intercepts.**

The parabola passes through the x- axis as (-2,0) and (3,0)

The solutions are x=-2 and x=3.

Solve the following equation by using a graph: 2x^2-3x+1.125.

**Graph the quadratic equation.**

**Identify the \textbf{x}- intercepts.**

The parabola passes through the x- axis as (0.75,0). Since it crosses at the vertex of the parabola, there is only 1 solution.

The solution is x=0.75.

**See also**: Quadratic graphs

In order to solve quadratic equations by factoring:

**Make sure that the equation is equal to \bf{0}.****Fully factor the quadratic equation**.**Set each expression equal to \bf{0}.****Solve for \textbf{x}.**

Solve x^{2}-8x+15=0 by factoring.

**Make sure that the equation is equal to \bf{0}. **

x^2-8 x+15=0 \checkmark

**Fully factor the quadratic equation**

(x-3)(x-5)=0

**Set each expression equal to \textbf{0}. **

x-3=0 \quad x-5=0

**Solve for \textbf{x}. **

The opposite of -3 is +3, so add 3 to both sides of the equation.

The opposite of -5 is +5, so add 5 to both sides of the equation.

You can see the real roots of the quadratic equation are where the quadratic graph crosses the x- axis.

Solve 2x^{2}+5x+3=0 by factoring.

**Make sure that the equation is equal to \bf{0}. **

2 x^2+5 x+3=0 \checkmark

**Fully factor the quadratic equation**

(2x+3)(x+1)=0

**Set each expression equal to \textbf{0}. **

2 x+3=0 \quad x+1=0

**Solve for \textbf{x}. **

The opposite of +3 is -3, so subtract 3 to both sides of the equation.

The opposite of \times 2 is \div 2, so divide by 2 on both sides of the equation.

The opposite of +1 is -1, so -1 to both sides of the equation.

You can see the real roots of the quadratic equation are where the quadratic graph crosses the x- axis.

In order to solve quadratic equations with the quadratic formula:

**Identify the value of \bf{\textbf{a}, \textbf{b}} and \textbf{c} in a quadratic equation.****Substitute these values into the quadratic formula.****Solve the equation with a \bf{+}, and then with a \bf{-}.**

Solve x^2-2 x-15=0 using the quadratic formula.

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

\begin{aligned}
& a=1 \\\\
& b=-2 \\\\
& c=-15
\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)(-15)}}{2(1)}
\end{aligned}

**Solve the equation with a \bf{+}, and then with a \bf{-}. **

\begin{aligned} & x=\cfrac{-(-2)+\sqrt{(-2)^2-4(1)(-15)}}{2(1)} \\\\ & x=\cfrac{2+\sqrt{4-(-60)}}{2} \\\\ & x=\cfrac{2+8}{2} \\\\ & x=5 \end{aligned} \quad \begin{aligned} & x=\cfrac{-(-2)-\sqrt{(-2)^2-4(1)(-15)}}{2(1)} \\\\ & x=\cfrac{2-\sqrt{4-(-60)}}{2} \\\\ & x=\cfrac{2-8}{2} \\\\ & x=-3 \end{aligned}

You can see the roots of the quadratic equation are where the quadratic graph crosses the x- axis.

You can also check that the solutions are correct by substituting them into the original equation.

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

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

First, simplify the equation so that it is in the form ax^2+b x+c=0, which leaves a quadratic expression on the left side of the equation and 0 on the right side of the equation.

-6 x^2+x=-2 \hspace{0.5cm} *Combine the like terms (2x-x)

-6 x^2+x=-2 \hspace{0.5cm} *Add 2 to both sides so that the equation equals 0

\hspace{0.8cm} +2 \hspace{0.4cm}+2

-6 x^2+x+2=0

**Substitute these values into the quadratic formula.**

\begin{aligned}
& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\
& x=\cfrac{-(1) \pm \sqrt{(1)^2-4(-6)(2)}}{2(-6)}
\end{aligned}

**Solve the equation with a \bf{+}, and then with a \bf{-}. **

\begin{aligned} & x=\cfrac{-(1)+\sqrt{(1)^2-4(-6)(2)}}{2(-6)} \\\\ & x=\cfrac{-1+\sqrt{1-(-48)}}{-12} \\\\ & x=\cfrac{-1+7}{-12} \\\\ & x=-\cfrac{1}{2} \end{aligned} \quad \begin{aligned} & x=\cfrac{-(1)-\sqrt{(1)^2-4(-6)(2)}}{2(-6)} \\\\ & x=\cfrac{-1-\sqrt{1-(-48)}}{-12} \\\\ & x=\cfrac{-1-7}{-12} \\\\ & x=\cfrac{2}{3} \end{aligned}

You can see the roots of the quadratic equation are where the quadratic graph crosses the x- axis.

You can also check that the solutions are correct by substituting them into the original equation.

- Let students use a graphing calculator to explore how changing the values for a, b and c in a quadratic equation changes the parabola.

- When students start using the quadratic formula, still encourage them to explain the meaning of their solutions. This way they do not forget what exactly they are solving for. One way to do this is to incorporate many real world examples of the quadratic equation that require students to explain the roots within the context of the problem.

**Thinking the order of the factored equation matters**When you multiply two values the order doesn’t matter.

For example,

2\times 3=3\times 2

It is the same with quadratics.

(x-6)(x+4) means (x-6)\times (x+4)

So, (x-6)(x+4)=0 is the same as (x+4)(x-6)=0.

**Forgetting to solve after factoring**

Don’t forget to set the factored expression equal to zero and solve it. Always check you have answered the question.

**Thinking that an equation that does not look like \bf{\textbf{ax}^2+\textbf{b x}+\textbf{c}=0} is not quadratic**

In a quadratic equation a≠ 0, but b and c can be 0.

For example,

Here is a quadratic equation that involves two square terms (perfect squares).

Notice that b = 0, because there is no term with just the variable x. However, it is still a quadratic equation and therefore can be solved.

\begin{array}{ll} \\4 x^2-16=0 & \\\\ (2 x+4)(2 x-4)=0 \end{array}

\begin{array}{ll} 2 x+4=0 \\\\ x=-2 \end{array} \quad \begin{array}{ll} 2 x-4=0 \\\\ x=2 \end{array}

**Thinking an equation that cannot be factored has no solutions**

If a quadratic equation cannot be factored, you can still solve it by using the quadratic formula. To work out the number of real solutions a quadratic equation has you can use the discriminant.

**Not converting to the form \bf{\textbf{ax}^{2}+\textbf{b x}+\textbf{c}=0} before trying to factor**

While it is possible to factor a quadratic equation without it in the standard form, this can be challenging. It is recommended to use the general form shown.

- Quadratic formula
- Solving quadratic equations by graphing
- Factoring quadratic equations

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

x=1, \quad y=1

x=-5, \quad x=5

x=2, \quad y=1

x=0, \quad x=2

The parabola passes through the x- axis as (0,0) and (2,0).

The solutions are x = 0 and x = 2.

2) Solve x^2+10 x+9 by using its graph below.

x=-5, \quad y=-13

x=-9, \quad x=-1

x=0, \quad x=-5

x=1, \quad y=8

The parabola passes through the x- axis as (-1,0) and (-9,0).

The solutions are x = -1 and x = -9.

3) Solve {x}^2+5x+6=0 by factoring.

x=3, \quad x=-2

x=3, \quad x=2

x=-3, \quad x=-2

x=-3, \quad x=2

{x}^2+5x+6=0 can be factored as (x+2)(x+3).

Setting each factor equal to zero and solving leads to the solutions.

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

4) Solve {x}^2-x-20=0 by factoring.

x=-5, \quad x=-4

x=5, \quad x=4

x=-5, \quad x=4

x=5, \quad x=-4

x^2-x-20 can be factored as (x-5)(x+4).

Setting each factor equal to zero and solving leads to the solutions.

\begin{aligned} & x-5=0 \\\\ & x=5 \end{aligned} \quad \begin{aligned} & x+4=0 \\\\ & x=-4 \end{aligned}

5) Solve 2{x}^2+3x-9=0 using the quadratic equation.

x=-\cfrac{3}{2}=-1.54 and x= -3

x=\cfrac{3}{2}=1.5 and x=3

x=-\cfrac{3}{2}=-1.5 and x=3

x=\cfrac{3}{2}=1.5 and x=-3

Identify a, b and c.

a=2, b=3, c=-9

Substitute these values into the quadratic formula.

\begin{aligned} & x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\ & x=\cfrac{-3 \pm \sqrt{(3)^2-4(2)(-9)}}{2(2)} \end{aligned}

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

\begin{aligned} & x=\cfrac{-3+\sqrt{(3)^2-4(2)(-9)}}{2(2)} \\\\ & x=\cfrac{-3+\sqrt{9-(-72)}}{4} \\\\ & x=\cfrac{-3+9}{4} \\\\ & x=\cfrac{3}{2}=1.5 \end{aligned} \quad \begin{aligned} & x=\cfrac{-3-\sqrt{(3)^2-4(2)(-9)}}{2(2)} \\\\ & x=\cfrac{-3-\sqrt{9-(-72)}}{4} \\\\ & x=\cfrac{-3-9}{4} \\\\ & x=-3 \end{aligned}

*Divide the numerator by the denominator to convert \cfrac{3}{2} to decimal form.

6) Solve 3{x}^2-9x+6=0 using the quadratic equation.

x=2, \quad x=1

x=-2, \quad x=-1

x=2, \quad x=-1

x=-2, \quad x=1

Identify a, b and c.

a=3, b=-9, 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{-(-9) \pm \sqrt{(-9)^2-4(3)(6)}}{2(3)} \end{aligned}

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

\begin{aligned} & x=\cfrac{-(-9)+\sqrt{(-9)^2-4(3)(6)}}{2(3)} \\\\ & x=\cfrac{9+\sqrt{81-72}}{6} \\\\ & x=\cfrac{9+3}{6} \\\\ & x=2 \end{aligned} \quad \begin{aligned} & x=\cfrac{-(-9)-\sqrt{(-9)^2-4(3)(6)}}{2(3)} \\\\ & x=\cfrac{9-\sqrt{81-72}}{6} \\\\ & x=\cfrac{9-3}{6} \\\\ & x=1 \end{aligned}

It is a polynomial equation whose highest variable is to the second-degree ( exponent of 2).

The standard form is ax^2+b x+c=0 or f(x)=a x^2+b x+c. The coefficients a, b and c can be whole numbers, integers, fractions, decimals or any other real number where a ≠ 0.

It is the value below the radical in the quadratic formula. If the discriminant b^2-4ac is positive, there are two solutions. If it is 0, there is 1 solution. If it is negative, there are no real solutions.

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