2.5: Solving Quadratics

Learning Objectives

• Solve quadratic equations by factoring.
• Solve quadratic equations by the square root property.
• Solve quadratic equations by completing the square.
• Solve quadratic equations by using the quadratic formula.

Solving Quadratic Equations by Factoring

An equation containing a second-degree polynomial is called a quadratic equation. For example, equations such as \(2x^2 +3x−1=0\) and \(x^2−4= 0\) are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.

Often the easiest method of solving a quadratic equation is factoring. Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation.

If a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property, which states that if \(a⋅b=0\), then \(a = 0\) or \(b =0\), where a and b are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero.

Multiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring. For example, expand the factored expression \((x−2)(x+3)\) by multiplying the two factors together.

\[\begin{align*} (x-2)(x+3)&= x^2+3x-2x-6\\ &= x^2+x-6\\ \end{align*}\]

The product is a quadratic expression. Set equal to zero, \(x^2+x−6= 0\) is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied.

The process of factoring a quadratic equation depends on the leading coefficient, whether it is \(1\) or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, \(ax^2+bx+c=0\), where \(a\), \(b\), and \(c\) are real numbers, and \(a≠0\). The equation \(x^2 +x−6= 0\) is in standard form.

We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor(GCF), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section.

ZERO-PRODUCT PROPERTY AND QUADRATIC EQUATIONS

The zero-product property states

If \(a⋅b=0\), then \(a=0\) or \(b=0\),

where \(a\) and \(b\) are real numbers or algebraic expressions.

A quadratic equation is an equation containing a second-degree polynomial; for example

\[ax^2+bx+c=0\]

where \(a\), \(b\), and \(c\) are real numbers, and if \(a≠0\), it is in standard form.

Solving Quadratics by Factoring

In the quadratic equation \(x^2 +x−6=0\), the leading coefficient, or the coefficient of \(x^2\), is \(1\). We begin by reviewing and practicing methods of factoring quadratic equations.

Example \(\PageIndex{1}\): Solving a Quadratic with Leading Coefficient of \(1\)

Factor and solve the equation: \(x^2+x−6=0\).

Solution

The two numbers that multioply to be \(-6\) and add to be \(1\) are \(3⋅(−2)\).  Then, write the factors.

\[(x−2)(x+3)=0 \nonumber \]

To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.

\[\begin{align*} (x-2)(x+3)&= 0\\ (x-2)&= 0\\ x&= 2\\ (x+3)&= 0\\ x&= -3 \end{align*}\]

The two solutions are \(2\) and \(−3\). We can see how the solutions relate to the graph in Figure \(\PageIndex{2}\). The solutions are the x-intercepts of \(x^2 +x−6=0\).

Figure \(\PageIndex{2}\)

Exercise \(\PageIndex{1}\): Solving a Quadratic with Leading Coefficient of \(1\) 

Solve the quadratic equation by factoring: \(x^2+8x+15=0\).

Answer

The numbers that add to \(8\) are \(3\) and \(5\). Then, write the factors, set each factor equal to zero, and solve.

\[\begin{align*} (x+3)(x+5)&= 0\\ (x+3)&= 0\\ x&= -3\\ (x+5)&= 0\\ x&= -5 \end{align*}\]

The solutions are \(−3\) and \(−5\).

Exercise \(\PageIndex{2}\): Solving a Quadratic with Leading Coefficient of \(1\)

Solve by factoring: \(x^2−9=0\).

Answer

Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the zero-factor property.

\[\begin{align*} x^2-9&= 0\\ (x-3)(x+3)&= 0\\ x-3&= 0\\ x&= 3\\ (x+3)&= 0\\ x&= -3 \end{align*}\]

The solutions are \(3\) and \(−3\).

When the leading coefficient is not \(1\), we factor a quadratic equation using the \(ac\) method, grouping, or grid method.  

Example \(\PageIndex{2}\): Solving a Quadratic Equation Using \(ac\) Method

Use grouping to factor and solve the quadratic equation: \(4x^2+15x+9=0\).

Solution

The only pair of factors that sums to \(15\) is \(3+12\). Rewrite the equation replacing the b term, \(15x\), with two terms using \(3\) and \(12\) as coefficients of \(x\). Factor the first two terms, and then factor the last two terms.

\[\begin{align*} 4x^2+3x+12x+9&= 0\\ x(4x+3)+3(4x+3)&= 0\\ (4x+3)(x+3)&= 0 \qquad \text{Solve using the zero-product property}\\ (4x+3)&= 3\\ x&= -\dfrac{3}{4}\\ (x+3)&= 0\\ x&= -3 \end{align*}\]

The solutions are \(−\dfrac{3}{4}\), and \(−3\). See Figure \(\PageIndex{3}\).

Figure \(\PageIndex{3}\)

Exercise \(\PageIndex{3}\)

Solve the equation by factoring: \(3x^2-x-10=0\).

Answer

Multiply \(3\) and \(10\).  Then determine which numbers multiply to be \(30\) and add to be \(-1\).  

\(3x^2-6x+5x-10=0\)

\(3x(x-2)+5(x-2)=0\)

\(3x+5)(x-2)=0\)

\(3x+5=0\) gives a solution of \(x=-\frac{5}{3}\)

\x-2=0\) gives a solution of \(x=2\)

The solutions are \(2\) and \(−\dfrac{5}{3}\).

Exercise \(\PageIndex{4}\):  Combining Skills

Solve the equation by factoring: \(−3x^3−5x^2−2x=0\).

Answer

This equation does not look like a quadratic, as the highest power is \(3\), not \(2\). Recall that the first thing we want to do when solving any equation is to factor out the GCF, if one exists. And it does here. We can factor out \(−x\) from all of the terms and then proceed with grouping.

\[\begin{align*}
-3x^3-5x^2-2x&= 0\\
-x(3x^2+5x+2)&= 0\\
-x(3x^2+3x+2x+2)&= 0 \qquad \text{Use grouping on the expression in parentheses}\\
-x[3x(x+1)+2(x+1)]&= 0\\
-x(3x+2)(x+1)&= 0\\
\text{Now, we use the zero-product property. Notice that we have three factors.}\\
-x&= 0\\
x&= 0\\
3x+2&= 0\\
x&= -\dfrac{2}{3}\\
x+1&= 0\\
x&= -1
\end{align*}\]

The solutions are \(0\), \(−\dfrac{2}{3}\), and \(−1\).

Using the Square Root Property

When there is no linear term in the equation, another method of solving a quadratic equation is by using the square root property, in which we isolate the \(x^2\) term and take the square root of the number on the other side of the equals sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the \(x^2\) term so that the square root property can be used.

THE SQUARE ROOT PROPERTY

With the \(x^2\) term isolated, the square root property states that:

if \(x^2=k\), then \(x=±\sqrt{k}\)

where \(k\) is a nonzero real number.

Example \(\PageIndex{3}\): Solving a Simple Quadratic Equation Using the Square Root Property

Solve the quadratic using the square root property: \(x^2=8\).

Solution

Take the square root of both sides, and then simplify the radical. Remember to use a \(±\) sign before the radical symbol.

\[\begin{align*} x^2&= 8\\ x&= \pm \sqrt{8}\\ &= \pm 2\sqrt{2} \end{align*}\]

The solutions are \(2\sqrt{2}\),\(-2\sqrt{2}\)

Exercise \(\PageIndex{5}\)

Solve the quadratic equation: \(4x^2+1=7\).

Answer

First, isolate the \(x^2\) term. Then take the square root of both sides.

\[\begin{align*} 4x^2+1&= 7\\ 4x^2&= 6\\ x^2&= \dfrac{6}{4}\\ x&= \pm \dfrac{\sqrt{6}}{2} \end{align*}\]

The solutions are \(\dfrac{\sqrt{6}}{2}\), and \(-\dfrac{\sqrt{6}}{2}\).

Completing the Square

Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, \(a\), must equal \(1\). If it does not, then divide the entire equation by \(a\). Then, we can use the following procedures to solve a quadratic equation by completing the square.

We will use the example \(x^2+10x+1=0\) to illustrate each step.

Given a quadratic equation that cannot be factored, and with \(a=1\), first add or subtract the constant term to the right sign of the equal sign.

\[\begin{align*}
x^2+10x+1&= 0\\
x^2+10x&= -1 \qquad \text{Multiply the b} \text{ term by } \dfrac{1}{2} \text{ and square it.}\\
\dfrac{1}{2}(10)&= 5 \\
5^2&= 25 \qquad \text{Add } \left ({\dfrac{1}{2}} \right )^2 \text{ to both sides of the equal sign and simplify the right side. We have}\\
x^2+10x+25&= -1+25\\
x^2+10x+25&= 24 \qquad \text{The left side of the equation can now be factored as a perfect square.}\\
{(x+5)}^2&=24\\
\sqrt{{(x+5)}^2}&= \pm \sqrt{24} \qquad \text{Use the square root property and solve.}\\
\sqrt{{(x+5)}^2}&= \pm \sqrt{24}\\
x+5&= \pm 2\sqrt{6}\\
x&= -5 \pm 2\sqrt{6}
\end{align*}\]

The solutions are \(−5+2\sqrt{6}\), and \(−5−2\sqrt{6}\).

Exercise \(\PageIndex{6}\):

Solve the quadratic equation by completing the square: \(x^2−3x−5=0\).

Answer

First, move the constant term to the right side of the equal sign.

\[\begin{align*}
x^2-3x&= 5 \qquad \text{Then, take } \dfrac{1}{2} \text{ of the b term and square it.} \\
\dfrac{1}{2}(-3)&= -\dfrac{3}{2}\\
{\left (-\dfrac{3}{2} \right )}^2=\dfrac{9}{4}\\
x^2-3x+{\left (-\dfrac{3}{2} \right )}^2&= 5+{\left (-\dfrac{3}{2} \right )}^2 \qquad \text{Add the result to both sides of the equal sign.}\\
x^2-3x+\dfrac{9}{4}&= 5+\dfrac{9}{4}\\
\text{Factor the left side as a perfect square and simplify the right side.}\\
{\left (x-\dfrac{3}{2} \right )}^2&= \dfrac{29}{4}\\
(x-\dfrac{3}{2})&= \pm \dfrac{\sqrt{29}}{2} \qquad \text{Use the square root property and solve.}\\
x&= \dfrac{3}{2} \pm \dfrac{\sqrt{29}}{2}\\
\end{align*}\]

The solutions are \(\dfrac{3}{2}+\dfrac{\sqrt{29}}{2}\), and \(\dfrac{3}{2}-\dfrac{\sqrt{29}}{2}\)

Using the Quadratic Formula

The fourth method of solving a quadratic equation is by using the quadratic formula, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.

We can derive the quadratic formula by completing the square. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by \(−1\) and obtain a positive a. Given \(ax^2+bx+c=0, a≠0\), we will complete the square as follows:

First, move the constant term to the right side of the equal sign:

\[ax^2+bx=−c \nonumber \]

As we want the leading coefficient to equal \(1\), divide through by \(a\):

\[x^2+\dfrac{b}{a}x=−\dfrac{c}{a} \nonumber \]

Then, find \(\dfrac{1}{2}\) of the middle term, and add \({(\dfrac{1}{2}\dfrac{b}{a})}^2=\dfrac{b^2}{4a^2}\) to both sides of the equal sign:

\[x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}=\dfrac{b^2}{4a^2}-\dfrac{c}{a} \nonumber \]

Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:

\[{(x+\dfrac{b}{2a})}^2=\dfrac{b^2-4ac}{4a^2} \nonumber \]

Now, use the square root property, which gives

\[x+\dfrac{b}{2a}=±\sqrt{\dfrac{b^2-4ac}{4a^2}} \nonumber \]

\[x+\dfrac{b}{2a}=\dfrac{±\sqrt{b^2-4ac}}{2a} \nonumber \]

Finally, add \(-\dfrac{b}{2a}\) to both sides of the equation and combine the terms on the right side. Thus,

\[x=\dfrac{-b±\sqrt{b^2-4ac}}{2a} \nonumber \]

THE QUADRATIC FORMULA

Written in standard form, \(ax^2+bx+c=0\), any quadratic equation can be solved using the quadratic formula:

\[x=\dfrac{-b±\sqrt{b^2-4ac}}{2a}\]

where \(a\), \(b\), and \(c\) are real numbers and \(a≠0\).

Example \(\PageIndex{4}\): Solve the Quadratic Equation Using the Quadratic Formula

Solve the quadratic equation: \(x^2+5x+1=0\).

Solution

Identify the coefficients: \(a=1,b=5,c=1\). Then use the quadratic formula.

\[\begin{align*} x&= \dfrac{-(5) \pm \sqrt{(5)^2-4(1)(1)}}{2(1)}\\ &= \dfrac{-5 \pm \sqrt{25-4}}{2}\\ &= \dfrac{-5 \pm \sqrt{21}}{2} \end{align*}\]

Exercise \(\PageIndex{7}\)

Use the quadratic formula to solve \(x^2+x+2=0\).

Answer

First, we identify the coefficients: \(a=1\),\(b=1\), and \(c=2\).

Substitute these values into the quadratic formula.

\[\begin{align*} x&= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\ &= \dfrac{-(1) \pm \sqrt{(1)^2-4(1)(2)}}{2(1)}\\ &= \dfrac{-1 \pm \sqrt{1-8}}{2}\\ &= \dfrac{-1 \pm \sqrt{-7}}{2}\\ &= \dfrac{-1 \pm i\sqrt{7}}{2} \end{align*}\]

The Discriminant

The quadratic formula not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the discriminant, or the expression under the radical, \(b^2−4ac\). The discriminant tells us whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. Table \(\PageIndex{1}\) relates the value of the discriminant to the solutions of a quadratic equation.

Table \(\PageIndex{1}\)

Value of Discriminant	Results
\(b^2−4ac=0\)	One rational solution (double solution)
\(b^2−4ac>0\), perfect square	Two rational solutions
\(b^2−4ac>0\), not a perfect square	Two irrational solutions
\(b^2−4ac<0\)	Two complex solutions

Example \(\PageIndex{5}\): Using the Discriminant to Find the Nature of the Solutions to a Quadratic Equation

Use the discriminant to find the nature of the solutions to the following quadratic equations:

1. \(x^2+4x+4=0\)
2. \(8x^2+14x+3=0\)
3. \(3x^2−5x−2=0\)
4. \(3x^2−10x+15=0\)

Solution

Calculate the discriminant \(b^2−4ac\) for each equation and state the expected type of solutions.

a.

\(x^2+4x+4=0\)

\(b^2-4ac={(4)}^2-4(1)(4)=0\) There will be one rational double solution.

b.

\(8x^2+14x+3=0\)

\(b^2-4ac={(14)}^2-4(8)(3)=100\) As \(100\) is a perfect square, there will be two rational solutions.

c.

\(3x^2−5x−2=0\)

\(b^2-4ac={(-5)}^2-4(3)(-2)=49\) As \(49\) is a perfect square, there will be two rational solutions.

d.

\(3x^2−10x+15=0\)

\(b^2-4ac={(-10)}^2-4(3)(15)=-80\) There will be two complex solutions.

Key Concepts

 

• Many quadratic equations can be solved by factoring when the equation has a leading coefficient of \(1\) or if the equation is a difference of squares. The zero-factor property is then used to find solutions. 
• Many quadratic equations with a leading coefficient other than \(1\) can be solved by factoring using the grouping method. 
• Another method for solving quadratics is the square root property. The variable is squared. We isolate the squared term and take the square root of both sides of the equation. The solution will yield a positive and negative solution. 
• Completing the square is a method of solving quadratic equations when the equation cannot be factored. 
• A highly dependable method for solving quadratic equations is the quadratic formula, based on the coefficients and the constant term in the equation. 
• The discriminant is used to indicate the nature of the roots that the quadratic equation will yield: real or complex, rational or irrational, and how many of each.