Section 2.1: Solving Quadratic Equations

Learning Outcomes

By the end of this section, you will be able to:

Factor a quadratic equation to solve it.
Use the square root property to solve a quadratic equation.
Use the Pythagorean Theorem and the square root property to find the unknown length of the side of a right triangle.
Complete the square to solve a quadratic equation.
Use the quadratic formula to solve a quadratic equation.
Use the discriminant to determine the number and type of solutions to a quadratic equation.

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Factoring and the Square Root Property

An equation containing a second-degree polynomial is called a quadratic equation. For example, equations such as $2{x}^{2}+3x - 1=0$ 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 $a\cdot b=0$ , 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 $\left(x - 2\right)\left(x+3\right)$ by multiplying the two factors together.

\begin{array}{l} (x - 2) (x + 3) & = x^{2} + 3 x - 2 x - 6 \\ = x^{2} + x - 6 \end{array}

$\begin{array}{l}\left(x - 2\right)\left(x+3\right)\hfill&={x}^{2}+3x - 2x - 6\hfill \\ \hfill&={x}^{2}+x - 6\hfill \end{array}$

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, $a{x}^{2}+bx+c=0$ , where a, b, and c are real numbers and $a\ne 0$ . 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.

A General Note: The Zero-Product Property and Quadratic Equations

The zero-product property states

If a \cdot b = 0, then a = 0 or b = 0

$\text{If }a\cdot b=0,\text{ then }a=0\text{ or }b=0$ ,

where a and b are real numbers or algebraic expressions.

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

a x^{2} + b x + c = 0

$a{x}^{2}+bx+c=0$

where a, b, and c are real numbers, and $a\ne 0$ . It is in standard form.

Solving Quadratics with a Leading Coefficient of 1

In the quadratic equation ${x}^{2}+x - 6=0$ , the leading coefficient, or the coefficient of ${x}^{2}$ , is 1. We have one method of factoring quadratic equations in this form.

How To: Given a quadratic equation with the leading coefficient of 1, factor it

Find two numbers whose product equals c and whose sum equals b.
Use those numbers to write two factors of the form $\left(x+k\right)\text{ or }\left(x-k\right)$ , where k is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and $-2$ , the factors are $\left(x+1\right)\left(x - 2\right)$ .
Solve using the zero-product property by setting each factor equal to zero and solving for the variable.

Example 1: Factoring and Solving a Quadratic with Leading Coefficient of 1

Factor and solve the equation: ${x}^{2}+x - 6=0$ .

Show Solution

To factor ${x}^{2}+x - 6=0$ , we look for two numbers whose product equals $-6$ and whose sum equals 1. Begin by looking at the possible factors of $-6$ .

\begin{array}{l} 1 \cdot (- 6) \\ (- 6) \cdot 1 \\ 2 \cdot (- 3) \\ 3 \cdot (- 2) \end{array}

$\begin{array}{l}1\cdot \left(-6\right)\hfill \\ \left(-6\right)\cdot 1\hfill \\ 2\cdot \left(-3\right)\hfill \\ 3\cdot \left(-2\right)\hfill \end{array}$

The last pair, $3\cdot \left(-2\right)$ sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.

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

$\left(x - 2\right)\left(x+3\right)=0$

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

\begin{array}{l} (x - 2) (x + 3) = 0 \\ (x - 2) = 0 \\ x = 2 \\ (x + 3) = 0 \\ x = - 3 \end{array}

$\begin{array}{l}\left(x - 2\right)\left(x+3\right)=0\hfill \\ \left(x - 2\right)=0\hfill \\ x=2\hfill \\ \left(x+3\right)=0\hfill \\ x=-3\hfill \end{array}$

The two solutions are $x=2$ and $x=-3$ . We can see how the solutions relate to the graph below. The solutions are the x-intercepts of the graph of ${x}^{2}+x - 6$ .

Coordinate plane with the x-axis ranging from negative 5 to 5 and the y-axis ranging from negative 7 to 7. The function x squared plus x minus six equals zero is graphed, with the x-intercepts (-3,0) and (2,0), plotted as well.

Try It

Factor and solve the quadratic equation: ${x}^{2}-5x - 6=0$ .

Show Solution

$\left(x - 6\right)\left(x+1\right)=0;x=6,x=-1$

Factoring by Grouping

Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial $2{x}^{2}+5x+3$ can be rewritten as $\left(2x+3\right)\left(x+1\right)$ using this process. We begin by rewriting the original expression as $2{x}^{2}+2x+3x+3$ and then factor each portion of the expression to obtain $2x\left(x+1\right)+3\left(x+1\right)$ . We then pull out the GCF of $\left(x+1\right)$ to find the factored expression.

A General Note: Factoring by Grouping

To factor a trinomial of the form $a{x}^{2}+bx+c$ by grouping, we find two numbers with a product of $ac$ and a sum of $b$ . We use these numbers to divide the $x$ term into the sum of two terms and factor each portion of the expression separately then factor out the GCF of the entire expression.

How To: Given a trinomial in the form $a{x}^{2}+bx+c$ , factor by grouping

List factors of $ac$ .
Find $p$ and $q$ , a pair of factors of $ac$ with a sum of $b$ .
Rewrite the original expression as $a{x}^{2}+px+qx+c$ .
Pull out the GCF of $a{x}^{2}+px$ .
Pull out the GCF of $qx+c$ .
Factor out the GCF of the expression.

Example 2: Factoring a Trinomial by Grouping

Factor $5{x}^{2}+7x - 6$ by grouping.

Show Solution

We have a trinomial with $a=5,b=7$ , and $c=-6$ . First, determine $ac=-30$ . We need to find two numbers with a product of $-30$ and a sum of $7$ . In the table, we list factors until we find a pair with the desired sum.

Factors of $-30$	Sum of Factors
$1,-30$	$-29$
$-1,30$	$29$
$2,-15$	$-13$
$-2,15$	$13$
$3,-10$	$-7$
$-3,10$	$7$

So $p=-3$ and $q=10$ .

\begin{array}{cc} 5 x^{2} - 3 x + 10 x - 6 & Rewrite the original expression as a x^{2} + p x + q x + c . \\ x (5 x - 3) + 2 (5 x - 3) & Factor out the GCF of each part . \\ (5 x - 3) (x + 2) & Factor out the GCF of the expression . \end{array}

$\begin{array}{cc}5{x}^{2}-3x+10x - 6 \hfill & \text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\hfill \\ x\left(5x - 3\right)+2\left(5x - 3\right)\hfill & \text{Factor out the GCF of each part}.\hfill \\ \left(5x - 3\right)\left(x+2\right)\hfill & \text{Factor out the GCF}\text{ }\text{ of the expression}.\hfill \end{array}$

Analysis of the Solution

We can check our work by multiplying. Use FOIL to confirm that $\left(5x - 3\right)\left(x+2\right)=5{x}^{2}+7x - 6$ .

Try It

Factor the following.

$2{x}^{2}+9x+9$
$6{x}^{2}+x - 1$

Show Solution

$\left(2x+3\right)\left(x+3\right)$
$\left(3x - 1\right)\left(2x+1\right)$

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 equal 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.

A General Note: The Square Root Property

With the ${x}^{2}$ term isolated, the square root propty states that:

if x^{2} = k, then x = \pm \sqrt{k}

$\text{if }{x}^{2}=k,\text{then }x=\pm \sqrt{k}$

where k is a nonzero real number.

How To: Given a quadratic equation with an ${x}^{2}$ term but no $x$ term, use the square root property to solve it

Isolate the ${x}^{2}$ term on one side of the equal sign.
Take the square root of both sides of the equation, putting a $\pm$ sign before the expression on the side opposite the squared term.
Simplify the numbers on the side with the $\pm$ sign.

Example 3: Solving a Simple Quadratic Equation Using the Square Root Property

Solve the quadratic using the square root property: ${x}^{2}=8$ .

Show Solution

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

\begin{array}{l} x^{2} & = 8 \\ x & = \pm \sqrt{8} \\ = \pm 2 \sqrt{2} \end{array}

$\begin{array}{l}{x}^{2}\hfill&=8\hfill \\ x\hfill&=\pm \sqrt{8}\hfill \\ \hfill&=\pm 2\sqrt{2}\hfill \end{array}$

The solutions are $x=2\sqrt{2}$ , $x=-2\sqrt{2}$ .

Example 4: Solving a Quadratic Equation Using the Square Root Property

Solve the quadratic equation: $\left(2x-1\right)^{2}-9=0$

Show Solution

First, isolate the $\left(2x-1\right)^{2}$ term. Then take the square root of both sides.

\begin{array}{l} {(2 x - 1)}^{2} - 9 = 0 \\ {(2 x - 1)}^{2} = 9 \\ 2 x - 1 = \pm 3 \\ x = \frac{\pm 3 + 1}{2} \\ x = - 1, 2 \end{array}

$\begin{array}{l}\left(2x-1\right)^{2}-9=0\hfill \\ \left(2x-1\right)^{2}=9\hfill \\ 2x-1 = \pm 3\hfill \\ x=\frac{\pm 3 + 1}{2}\hfill \\ x=-1,2 \end{array}$

The solutions are $x=-1$ , $x=2$ .

Try It

Solve the quadratic equation using the square root property: $3{\left(x - 4\right)}^{2}=15$ .

Show Solution

$x=4\pm \sqrt{5}$

Using the Pythagorean Theorem

One of the most famous formulas in mathematics is the Pythagorean Theorem. It is based on a right triangle and states the relationship among the lengths of the sides as ${a}^{2}+{b}^{2}={c}^{2}$ , where $a$ and $b$ refer to the legs of a right triangle adjacent to the $90^\circ$ angle, and $c$ refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications.

We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side.

The Pythagorean Theorem is given as

a^{2} + b^{2} = c^{2}

${a}^{2}+{b}^{2}={c}^{2}$

where $a$ and $b$ refer to the legs of a right triangle adjacent to the ${90}^{\circ }$ angle, and $c$ refers to the hypotenuse.

Right triangle with the base labeled: a, the height labeled: b, and the hypotenuse labeled: c

Example 5: Finding the Length of the Missing Side of a Right Triangle

Find the length of the missing side of the right triangle.

Right triangle with the base labeled: a, the height labeled: 4, and the hypotenuse labeled 12.

Show Solution

As we have measurements for side b and the hypotenuse, the missing side is a.

\begin{array}{l} a^{2} + b^{2} = c^{2} \\ a^{2} + {(4)}^{2} = {(12)}^{2} \\ a^{2} + 16 = 144 \\ a^{2} = 128 \\ a = \sqrt{128} \\ a = 8 \sqrt{2} \end{array}

$\begin{array}{l}{a}^{2}+{b}^{2}={c}^{2}\hfill \\ {a}^{2}+{\left(4\right)}^{2}={\left(12\right)}^{2}\hfill \\ {a}^{2}+16=144\hfill \\ {a}^{2}=128\hfill \\ a=\sqrt{128}\hfill \\ a=8\sqrt{2}\hfill \end{array}$

Try It

Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse.

Show Solution

$5$ units

Completing the Square and the Quadratic Formula

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 ${x}^{2}+4x+1=0$ 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.
${x}^{2}+4x=-1$
Multiply the b term by $\frac{1}{2}$ and square it.
$\begin{array}{l}\frac{1}{2}\left(4\right)=2\hfill \\ {2}^{2}=4\hfill \end{array}$
Add ${\left(\frac{1}{2}b\right)}^{2}$ to both sides of the equal sign and simplify the right side. We have
$\begin{array}{l}{x}^{2}+4x+4=-1+4\hfill \\ {x}^{2}+4x+4=3\hfill \end{array}$
The left side of the equation can now be factored as a perfect square.
$\begin{array}{l}{x}^{2}+4x+4=3\hfill \\ {\left(x+2\right)}^{2}=3\hfill \end{array}$
Use the square root property and solve.
$\begin{array}{l}\sqrt{{\left(x+2\right)}^{2}}=\pm \sqrt{3}\hfill \\ x+2=\pm \sqrt{3}\hfill \\ x=-2\pm \sqrt{3}\hfill \end{array}$
The solutions are $x=-2+\sqrt{3}$ , $x=-2-\sqrt{3}$ .

Example 6: Solving a Quadratic by Completing the Square

Solve the quadratic equation by completing the square: ${x}^{2}-3x - 5=0$ .

Show Solution

First, move the constant term to the right side of the equal sign by adding 5 to both sides of the equation.

x^{2} - 3 x = 5

${x}^{2}-3x=5$

Then, take $\frac{1}{2}$ of the b term and square it.

\begin{array}{l} \frac{1}{2} (- 3) = - \frac{3}{2} \\ {(- \frac{3}{2})}^{2} = \frac{9}{4} \end{array}

$\begin{array}{l}\frac{1}{2}\left(-3\right)=-\frac{3}{2}\hfill \\ {\left(-\frac{3}{2}\right)}^{2}=\frac{9}{4}\hfill \end{array}$

Add the result to both sides of the equal sign.

\begin{array}{l} x^{2} - 3 x + {(- \frac{3}{2})}^{2} = 5 + {(- \frac{3}{2})}^{2} \\ x^{2} - 3 x + \frac{9}{4} = 5 + \frac{9}{4} \end{array}

$\begin{array}{l}\text{ }{x}^{2}-3x+{\left(-\frac{3}{2}\right)}^{2}=5+{\left(-\frac{3}{2}\right)}^{2}\hfill \\ {x}^{2}-3x+\frac{9}{4}=5+\frac{9}{4}\hfill \end{array}$

Factor the left side as a perfect square and simplify the right side.

{(x - \frac{3}{2})}^{2} = \frac{29}{4}

${\left(x-\frac{3}{2}\right)}^{2}=\frac{29}{4}$

Use the square root property and solve.

$\begin{array}{c}\sqrt{{\left(x-\frac{3}{2}\right)}^{2}}=\pm \sqrt{\frac{29}{4}}\hfill \\ \left(x-\frac{3}{2}\right)=\pm \frac{\sqrt{29}}{2}\hfill \\ x=\frac{3}{2}\pm \frac{\sqrt{29}}{2}\hfill \end{array}$

The solutions are $x=\frac{3}{2}+\frac{\sqrt{29}}{2}$ , $x=\frac{3}{2}-\frac{\sqrt{29}}{2}$ .

Try It

Solve by completing the square: ${x}^{2}-6x=13$ .

Show Solution

$x=3\pm \sqrt{22}$

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 $a{x}^{2}+bx+c=0$ , $a\ne 0$ , we will complete the square as follows:

First, move the constant term to the right side of the equal sign:
$a{x}^{2}+bx=-c$
As we want the leading coefficient to equal 1, divide through by a:
${x}^{2}+\frac{b}{a}x=-\frac{c}{a}$
Then, find $\frac{1}{2}$ of the middle term, and add ${\left(\frac{1}{2}\frac{b}{a}\right)}^{2}=\frac{{b}^{2}}{4{a}^{2}}$ to both sides of the equal sign:
${x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}}=\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}$
Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:
${\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}$
Now, use the square root property, which gives
$\begin{array}{l}x+\frac{b}{2a}=\pm \sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\hfill \\ x+\frac{b}{2a}=\frac{\pm \sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}$
Finally, add $-\frac{b}{2a}$ to both sides of the equation and combine the terms on the right side. Thus,
$x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$

A General Note: The Quadratic Formula

Written in standard form, $a{x}^{2}+bx+c=0$ , any quadratic equation can be solved using the quadratic formula:

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

$x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$

where a, b, and c are real numbers and $a\ne 0$ .

How To: Given a quadratic equation, solve it using the quadratic formula

Make sure the equation is in standard form: $a{x}^{2}+bx+c=0$ .
Make note of the values of the coefficients and constant term, $a,b$ , and $c$ .
Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.
Calculate and solve.

Example 7: Solve the Quadratic Equation Using the Quadratic Formula

Solve the quadratic equation: ${x}^{2}+5x+1=0$ .

Show Solution

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

\begin{array}{l} x & = \frac{- (5) \pm \sqrt{{(5)}^{2} - 4 (1) (1)}}{2 (1)} \\ = \frac{- 5 \pm \sqrt{25 - 4}}{2} \\ = \frac{- 5 \pm \sqrt{21}}{2} \end{array}

$\begin{array}{l}x\hfill&=\frac{-\left(5\right)\pm \sqrt{{\left(5\right)}^{2}-4\left(1\right)\left(1\right)}}{2\left(1\right)}\hfill \\ \hfill&=\frac{-5\pm \sqrt{25 - 4}}{2}\hfill \\ \hfill&=\frac{-5\pm \sqrt{21}}{2}\hfill \end{array}$

Example 8: Solving a Quadratic Equation with the Quadratic Formula

Use the quadratic formula to solve ${x}^{2}+x+2=0$ .

Show Solution

First, we identify the coefficients: $a=1,b=1$ , and $c=2$ .

Substitute these values into the quadratic formula.

\begin{array}{l} x & = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = \frac{- (1) \pm \sqrt{{(1)}^{2} - (4) \cdot (1) \cdot (2)}}{2 \cdot 1} \\ = \frac{- 1 \pm \sqrt{1 - 8}}{2} \\ = \frac{- 1 \pm \sqrt{- 7}}{2} \\ = \frac{- 1 \pm i \sqrt{7}}{2} \end{array}

$\begin{array}{l}x\hfill&=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}\hfill \\\hfill&=\frac{-\left(1\right)\pm \sqrt{{\left(1\right)}^{2}-\left(4\right)\cdot \left(1\right)\cdot \left(2\right)}}{2\cdot 1}\hfill \\\hfill&=\frac{-1\pm \sqrt{1 - 8}}{2}\hfill \\ \hfill&=\frac{-1\pm \sqrt{-7}}{2}\hfill \\\hfill&=\frac{-1\pm i\sqrt{7}}{2}\hfill \end{array}$

The solutions to the equation are $x=\frac{-1}{2}+\frac{i\sqrt{7}}{2}$ and $x=\frac{-1}{2}-\frac{i\sqrt{7}}{2}$ .

Notice they are written in standard form of a complex number. When a solution is a complex number, you must separate the real part from the imaginary part and write it in standard form.

Try It

Solve the quadratic equation using the quadratic formula: $9{x}^{2}+3x - 2=0$ .

Show Solution

$x=-\frac{2}{3}$ , $x=\frac{1}{3}$

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 as well as how many solutions of each type to expect. The table below relates the value of the discriminant to the solutions of a quadratic equation.

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

A General Note: The Discriminant

For $a{x}^{2}+bx+c=0$ , where $a$ , $b$ , and $c$ are real numbers, the discriminant is the expression under the radical in the quadratic formula: ${b}^{2}-4ac$ . It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect.

Example 9: 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:

${x}^{2}+4x+4=0$
$8{x}^{2}+14x+3=0$
$3{x}^{2}-5x - 2=0$
$3{x}^{2}-10x+15=0$

Show Solution

Calculate the discriminant ${b}^{2}-4ac$ for each equation and state the expected type of solutions.

${x}^{2}+4x+4=0$ : ${b}^{2}-4ac={\left(4\right)}^{2}-4\left(1\right)\left(4\right)=0$ . There will be one rational double solution.
$8{x}^{2}+14x+3=0$ : ${b}^{2}-4ac={\left(14\right)}^{2}-4\left(8\right)\left(3\right)=100$ . As $100$ is a perfect square, there will be two rational solutions.
$3{x}^{2}-5x - 2=0$ : ${b}^{2}-4ac={\left(-5\right)}^{2}-4\left(3\right)\left(-2\right)=49$ . As $49$ is a perfect square, there will be two rational solutions.
$3{x}^{2}-10x+15=0$ : ${b}^{2}-4ac={\left(-10\right)}^{2}-4\left(3\right)\left(15\right)=-80$ . There will be two complex solutions.

Try It

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 solutions that the quadratic equation will yield: real or complex, rational or irrational, and how many of each.
The Pythagorean Theorem, among the most famous theorems in history, is used to solve right-triangle problems and has applications in numerous fields. Solving for the length of one side of a right triangle requires solving a quadratic equation.

Glossary

completing the square: a process for solving quadratic equations in which terms are added to or subtracted from both sides of the equation in order to make one side a perfect square

discriminant: the expression under the radical in the quadratic formula that indicates the nature of the solutions, real or complex, rational or irrational, single or double roots.

Pythagorean Theorem: a theorem that states the relationship among the lengths of the sides of a right triangle, used to solve right triangle problems

quadratic equation: an equation containing a second-degree polynomial; can be solved using multiple methods

quadratic formula: a formula that will solve all quadratic equations

square root property: one of the methods used to solve a quadratic equation in which the ${x}^{2}$ term is isolated so that the square root of both sides of the equation can be taken to solve for x

zero-product property: the property that formally states that multiplication by zero is zero so that each factor of a quadratic equation can be set equal to zero to solve equations

Section 2.1 Homework Exercises

1. How do we recognize when an equation is quadratic?

2. When we solve a quadratic equation, how many solutions should we always start seeking out? Explain why when solving a quadratic equation in the form $ax^{2}+bx+c=0$ we may graph the equation $\text{ }y=ax^{2}+bx+c$ and have no zeroes ( $x$ -intercepts).

3. When we solve a quadratic equation by factoring, why do we move all terms to one side, having zero on the other side?

4. In the quadratic formula, what is the name of the expression under the radical sign $\text{ }b^{2}-4ac$ , and how does it determine the number of and nature of our solutions?

5. Describe two scenarios where using the square root property to solve a quadratic equation would be the most efficient method.

For the following exercises, solve the quadratic equation by factoring.

6. $\text{ }x^{2}-4x-21=0$

7. $\text{ }x^{2}-9x+18=0$

8. $\text{ }2x^{2}+9x-5=0$

9. $\text{ }6x^{2}+17x+5=0$

10. $\text{ }4x^{2}-12x+8=0$

11. $\text{ }3x^{2}-75=0$

12. $\text{ }8x^{2}+6x-9=0$

13. $\text{ }4x^{2}=9$

14. $\text{ }2x^{2}+14x=36$

15. $\text{ }5x^{2}=5x+30$

16. $\text{ }4x^{2}=5x$

17. $\text{ }7x^{2}+3x=0$

18. $\text{ }\dfrac{x}{3}-\dfrac{9}{x}=2$

For the following exercises, solve the quadratic equation by using the square root property.

19. $\text{ }x^{2}=36$

20. $\text{ }x^{2}=49$

21. $\text{ }(x-1)^{2}=25$

22. $\text{ }(x-3)^{2}=7$

23. $\text{ }(2x+1)^{2}=9$

24. $\text{ }(x-5)^{2}=4$

For the following exercises, solve the quadratic equation by completing the square. Show each step.

25. $\text{ }x^{2}-9x-22=0$

26. $\text{ }2x^{2}-8x-5=0$

27. $\text{ }x^{2}-6x=13$

28. $\text{ }x^{2}+\dfrac{2}{3}x-\dfrac{1}{3}=0$

29. $\text{ }2+z=6z^{2}$

30. $\text{ }6p^{2}+7p-20=0$

31. $\text{ }2x^{2}-3x-1=0$

For the following exercises, determine the discriminant, and then state how many solutions there are and the nature of the solutions. Do not solve.

32. $\text{ }2x^{2}-6x+7=0$

33. $\text{ }x^{2}+4x+7=0$

34. $\text{ }3x^{2}+5x-8=0$

35. $\text{ }9x^{2}-30x+25=0$

36. $\text{ }2x^{2}-3x-7=0$

37. $\text{ }6x^{2}-x-2=0$

For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No Real Solution.

38. $\text{ }2x^{2}+5x+3=0$

39. $\text{ }x^{2}+x=4$

40. $\text{ }2x^{2}-8x-5=0$

41. $\text{ }3x^{2}-5x+1=0$

42. $\text{ }x^{2}+4x+2=0$

43. $\text{ }4+\dfrac{1}{x}-\dfrac{1}{x^{2}}=0$

44. $\text{ }x+\dfrac{4}{x}+6=0$

45. Beginning with the general form of the quadratic equation $\text{ }ax^{2}+bx+c=0$ , solve for x by using the completing the square method, thus deriving the quadratic formula.

46. Show that the sum of the two solutions to the quadratic formula is $-\dfrac{b}{a}$ .

47. A person has a garden that has a length 10 feet longer than the width. Set up a quadratic equation to find the dimensions of the garden if its area is $119 ft^{2}$ . Solve the quadratic equation to find the length and width.

48. Abercrombie and Fitch stock had a price given as $P=0.2t^{2}-5.6t+50.2$ , where $t$ is the time in months from 1999 to 2001. ( $t=1$ is January 1999). Find the two months in which the price of the stock was $30.

49. Suppose that an equation is given $p=-2x^{2}+280x-1000$ , where $x$ represents the number of items sold at an auction and $p$ is the profit made by the business that ran the auction. How many items solve would make this profit a maximum? Solve this by graphing the expression in your graphing utility and finding the maximum using 2nd CALC maximum. To obtain a good window for the curve, set $x$ [0,200] and $y$ [0,10000].

50. A formula for the normal systolic blood pressure for a man age A, measured in mmHg, is given as $P=0.006A^{2}-0.02A+120$ . Find the age to the nearest year of a man whose normal blood pressure measures 125 mmHg.

51. The cost function for a certain company is $C=60x+300$ and the revenue is given by $R=100x-0.5x^{2}$ . Recall the the profit is revenue minus cost. Set up a quadratic equation and find two values of $x$ (production level) that will create a profit of $300.

52. A falling object travels a distance given by the formula $d=5t+16t^{2}$ ft, where $t$ is measured in seconds. How long will it take for the object to travel 74 ft?

53. A vacanot lot is being converted into a community garden. The garden and walkway around its perimeter have an area of $378\text{ }ft^{2}$ . Find the width of the walkway if the garden is $12ft$ wide by $15ft$ long.
Garden with walkway border.

54. An epidemiological study of the spread of a certain influenza strain that hit a small school population found that the total number of students, P, who contracted the flu $t$ days after it broke out is given by the model $P=-t^{2}+13t+130$ , where $1<=t<=6$ . Find the day that 160 students had the flu. Recall that the restriction on $t$ is at most 6.

Chapter 2: Polynomial and Rational Functions