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.

## What is solving quadratic equations?

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}

## Common Core State Standards

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.

## How to solve a quadratic equation by graphing

In order to solve quadratic equations by graphing:

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

### Example 1: solve a quadratic equation by looking at its graph

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

2Identify 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.

### Example 2: solve a quadratic equation by looking at its graph

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

Identify the \textbf{x}- intercepts.

## How to solve a quadratic equation by factoring

In order to solve quadratic equations by factoring:

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

### Example 3: solve a quadratic equation by factoring with coefficient a=1

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

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

Set each expression equal to \textbf{0}.

Solve for \textbf{x}.

### Example 4: solve a quadratic equation by factoring with coefficient a>1

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

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

Set each expression equal to \textbf{0}.

Solve for \textbf{x}.

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

### Example 5: equation of the form ax²+b x+c=0 where a = 1

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.

Substitute these values into the quadratic formula.

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

### Example 6: solutions of the quadratic equation with a<1

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.

Substitute these values into the quadratic formula.

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

### Teaching tips for solving quadratic equations

• 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.

### Easy mistakes to make

• 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.

• Solving quadratic equations by graphing

### Solving quadratic equations questions (by factoring)

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

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.

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}^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^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.

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).

What is the standard form of a quadratic equation or a quadratic function?

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.

What is the discriminant?

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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