Solving Quadratic Equations

Concept

Quadratic Equations

In standard form, quadratic equations take the form $y=ax^2+bx+c,$ and can be solved in various ways. In general, they are solved for the value(s) of $x$ that make the equation equal to $0.$ Thus, $y=ax^2+bx+c$ becomes $0=ax^2+bx+c.$ Graphically, all points with a $y$ -coordinate of $0$ are the $x$ -intercepts of the function — or the zeros of the parabola. That means, solving a quadratic equation leads to finding the zeros of the parabola. Since a parabola can have $0, 1,$ or $2$ zeros, a quadratic equation can have $0, 1,$ or $2$ solutions.

Method

Solving a Quadratic Equation Graphically

A quadratic equation can be solved graphically by expressing the equation as a function and identifying the solutions in the function's graph. Consider the following equation as an example.

2x^2+3x-5=5x+7

This and other quadratic equations can be solved using this method.

1

If needed, rearrange the equation and simplify

If neither side of the equation equals $0,$ rearrange it so that all terms are on the same side. $2x^2+3x-5=5x+7 \quad \Leftrightarrow \quad 2x^2+3x-5-5x-7=0$ If there is more than one term of the same degree, simplify the equation by combining like terms. $2x^2+3x-5-5x-7=0 \quad \Leftrightarrow \quad 2x^2-2x-12=0$

2

Plot the function

The side with the variables can now be seen as a function, $f(x).$ Plot that function in a coordinate plane. For the equation $2x^2-2x-12=0,$ the function is $f(x) = 2x^2-2x-12.$

3

Identify any points with

y

-coordinate

0

Now, find all the points on the graph that have the $y$ -coordinate $0.$

4

Identify the

x

-coordinates

The $x$ -coordinates of any identified points solve the equation, $f(x)=0.$

In the example, the $x$ -coordinates are $\text{-} 2$ and $3.$ Note that solving an equation graphically does not necessarily lead to an exact answer. To verify a solution, substitute it into $f(x)$ and evaluate the expression. Verifying the solution $x = \text{-} 2$ is done by evaluating $f(\text{-} 2).$

f(x) = 2x^2-2x-12

x={\color{#0000FF}{\text{-} 2}}

f({\color{#0000FF}{\text{-} 2}}) = 2\cdot ({\color{#0000FF}{\text{-} 2}})^2-2\cdot({\color{#0000FF}{\text{-} 2}})-12

Calculate power

f(\text{-} 2) = 2\cdot 4-2\cdot(\text{-} 2)-12

\text{-} a(\text{-} b)=a\cdot b

f(\text{-} 2) = 2\cdot 4+2\cdot2-12

Multiply

f(\text{-} 2) = 8+4-12

Add and subtract terms

f(\text{-} 2) = 0

We find that the function value is

0.

Therefore,

x=\text{-} 2

is an exact solution. The solution

x = 3

can now be verified the same way.

f(x) = 2x^2-2x-12

x={\color{#0000FF}{3}}

f({\color{#0000FF}{3}}) = 2\cdot {\color{#0000FF}{3}}^2-2\cdot{\color{#0000FF}{3}}-12

Calculate power

f(3) = 2\cdot 9-2\cdot3-12

Multiply

f(3) = 18-6-12

Add and subtract terms

f(3) = 0

Since

f(3)=0

then

x=3

is also a solution. Thus, the equation

2x^2+3x-5=5x+7

has two solutions and they are

x=\text{-} 2 \quad \text{and} \quad x=3.

Concept

Simple Quadratic Equations

Simple quadratic equations take the form $ax^2+c=0$ and are solved using inverse operations. Once $x^2$ remains on the left-hand side, the equation can be written as $x^2=d,$ where $d=\frac{\text{-} c}{a}$ . The value of $d$ gives the number of solutions the equation has.

\begin{aligned} \mathbf{d>0} \quad &: \quad 2\text{ real solutions}\\ \mathbf{d=0} \quad &: \quad 1\text{ real solution}\\ \mathbf{d<0} \quad &: \quad 0\text{ real solutions} \end{aligned}

Method

Solving Quadratic Equations with Square Roots

Quadratic equations of the form

ax^2+c=0

can be solved using inverse operations. To undo the exponent of

2

on the

x^2

term, square roots must be used. Consider

5x^2-500=0.

1

Isolate

x^2

To begin, isolate

x^2

using inverse operations to move

c

and then

a

to the opposite side of the equation. Here, that means adding

500

and then dividing by

5.

5x^2-500=0

\text{LHS}+500=\text{RHS}+500

5x^2=500

\left.\text{LHS}\middle/5\right.=\left.\text{RHS}\middle/5\right.

x^2=100

2

Square-root both sides of the equation

Now that

x^2

has been isolated, it is necessary to undo the exponent. Exponents and radicals of the same index undo each other. Thus, square roots undo powers of

2

and cube roots undo powers of

3,

etc. To finish isolating

x^2,

square-root both sides of the equation.

x^2=100

\sqrt{\text{LHS}}=\sqrt{\text{RHS}}

x=\pm \sqrt{100}

Using a calculator to determine

\sqrt{100}

gives one value —

10.

This is because calculators give the principal root of a square root. It is necessary to remember that there are two values that, when raised to the power of

2,

equal

100.

10 \cdot 10 = 100 \quad \text{and} \quad (\text{-} 10)(\text{-} 10)=100,

Thus,

x=10

and

x=\text{-} 10.

Method

Factoring a Quadratic Trinomial

When factoring an expression in the form

ax^2+bx+c

it can be difficult to see its factors.

\begin{gathered} 4x^2+13x+3 \end{gathered}

This expression can be factored by finding a pair of integers whose product is

a\cdot c,

which here is

4\cdot 3,

and whose sum is

b,

which in the example is

13.

1

List all pairs of integers whose product is equal to

a\cdot c

The first step is to find all possible pairs of integers that multiply to $a\cdot c.$ In this case, $a=4$ and $c=3.$ Thus, their product is $3\cdot 4 = 12.$ To find all factor pairs, start with the pair where one factor is $1.$ The other factor must then be $12.$ Then continue with the pair where one of the factors is $2,$ and so forth. In this case, there are three pairs. $\begin{aligned} &1 \quad \text{and} \quad 12\\ &2 \quad \text{and} \quad 6\\ &3 \quad \text{and} \quad 4 \end{aligned}$

2

Find the pair whose sum is

b

If the given expression is factorable, one of the factor pairs will add to equal $b.$ In this case, $b=13.$ $\begin{aligned} 1+12&=13 \\ 2+6&=8 \\ 3+4&=7 \end{aligned}$ Here, $1$ and $12$ is the only factor pair that adds to $13.$

3

Write

bx

as a sum

Now, use the factor pair to rewrite the $x$ -term of the original expression as a sum. Since the factor pair is $1$ and $12,$ the middle term can be written as $13x = 12x + x.$ This gives the following equivalent expression. $\begin{aligned} 4x^2 &+ \ \ \quad {\color{#0000FF}{13x}}\ &&+ 3 \\ 4x^2 &+ \ \ \ {\color{#0000FF}{12x +x}} &&+ 3 \end{aligned}$ Notice that the expression hasn't been changed. Rather, it has been rewritten.

4

Factor out the GCF in the first two terms

Now, the expression has four terms, which can be grouped into the first two terms and the last two terms. Then, the GCF of each group can be factored out. Here, begin with the first group of terms, $4x^2$ and $12x.$ Notice how each can be written as a product of its factors. $\begin{aligned} 4x^2 \ &+ \ 12x \\ 4 \cdot x \cdot x \ &+ \ 4 \cdot 3 \cdot x \end{aligned}$ The GCF is $4x.$ Therefore, it's possible to factor out $4x.$ $\begin{aligned} 4x^2 +12x \textcolor{lightgray}{+}\textcolor{lightgray}{x} \textcolor{lightgray}{+} \textcolor{lightgray}{3} \\ 4x(x+3)\textcolor{lightgray}{+} \textcolor{lightgray}{x}\textcolor{lightgray}{+}\textcolor{lightgray}{3} \end{aligned}$

5

Factor out the GCF in the last two terms

Next, repeat the same process with the last two terms. In this case, $x$ and $3$ do not have any common factors, but it's always possible to write expressions as a product of $1$ and itself. $\begin{aligned} \textcolor{lightgray}{4x(x+3)} & \textcolor{lightgray}{+} x+3 \\ \textcolor{lightgray}{4x(x+3)} & \textcolor{lightgray}{+} 1(x+3) \end{aligned}$

6

Factor out the common factor

If all previous steps have been performed correctly, there should now be two terms with a common factor, which can be seen as a repeated parentheses. Factoring $(x+3)$ out of both terms gives $\begin{aligned} {\color{#0000FF}{4x}}(x+3) + & {\color{#0000FF}{1}} (x+3) \\ (x+3) ({\color{#0000FF}{4}}&{\color{#0000FF}{x}}+{\color{#0000FF}{1}}). \end{aligned}$ This means that $4x^2+13x+3$ can be written in factored form as $(x+3)(4x+1).$

Solve the quadratic equation by factoring

Exercise

Solve the following equation by factoring. $x^2-2x-3=0$

Solution

We'll focus on the left-hand side of the equation before solving for

x.

Notice that

1

and

\text{-} 3

are the factors that multiply to equal

\text{-} 3

and add to

\text{-} 2.

x^2 - 2x-3 = 0

Rewrite

\text{-} 2x

as

x-3x

x^2+x - 3x-3 = 0

Factor out

x

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

Factor out

\text{-} 3

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

Factor out

(x+1)

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

Notice that, in factored form, the numbers in the parentheses with

x

are the numbers from the chosen factor pair. This is because the coefficient of

x^2

is

1.

We could have written the expression in factored form immediately after finding the factor pair. Lastly, we can use the zero product property to solve the equation.

\begin{aligned} x+1=0 \quad &\leftrightarrow \quad x=\text{-} 1\\ x-3=0 \quad &\leftrightarrow \quad x= \phantom{\text{-}} 3 \end{aligned}

Thus, the solutions to the equation

x^2-2x-3=0

are

x=\text{-} 1

and

x=3.

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Formulary

Course 1

Course 2

Course 3

Course 4

Course 5

Solving Quadratic Equations

Quadratic Equations

Solving a Quadratic Equation Graphically

1

2

3

4

Simple Quadratic Equations

Solving Quadratic Equations with Square Roots

1

2

Factoring a Quadratic Trinomial

1

2

3

4

5

6

Example

Solve the quadratic equation by factoring