### Learning Objectives

• (10.1.1) – Use the zero product principle to solve quadratic equations that can be factored
• (10.1.2) – Solve a Quadratic Equation by the Square Root Property
• (10.1.3) – Solve a Quadratic Equation by Completing the Square
• (10.1.4) – The Quadratic Formula
• Use the Quadratic Formula to find solutions of a quadratic equation, (rational, irrational and complex)
• (10.1.5) – Use the discriminant to classify solutions of quadratic equations

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.

# (10.1.1) – Use the zero product principle to solve quadratic equations that can be factored

Often the easiest method of solving a quadratic equation is by factoring. Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation. Note that we will not spend a lot of time explaining how to factor in this section.  You may want to seek help if you don’t feel confident about factoring.

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.

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.

### The Zero-Product Property and Quadratic Equations

The zero-product property states

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

where $a$ and $b$ are real numbers or algebraic expressions.

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

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

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

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.

### Reminder: Given a quadratic equation with the leading coefficient of 1, factor it.

1. Find two numbers whose product equals $c$ and whose sum equals $b$.
2. 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 $\left(x+1\right)\left(x - 2\right)$.
3. Solve using the zero-product property by setting each factor equal to zero and solving for the variable.

### Example

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

In the following video we provide more examples of factoring to solve quadratic equations.

In our next example, we will solve a quadratic equation that is written as a difference of squares.

### Example

Solve the difference of squares equation using the zero-product property: ${x}^{2}-9=0$.

#### Solving Quadratics with a Leading Coefficient of $\ne1$

Recall that when the leading coefficient is not 1, we factor a quadratic equation using the method called grouping, which requires four terms. With the equation in standard form, let’s review the grouping procedures:

1. With the quadratic in standard form, $a{x}^{2}+bx+c=0$, multiply $a\cdot c$.
2. Find two numbers whose product equals $ac$ and whose sum equals $b$.
3. Rewrite the equation replacing the $bx$ term with two terms using the numbers found in step 1 as coefficients of $x$.
4. Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping.
5. Factor out the expression in parentheses.
6. Set the expressions equal to zero and solve for the variable.

### Example

Use grouping to factor and solve the quadratic equation: $4{x}^{2}+15x+9=0$.

The following video contains another example of solving a quadratic equation using factoring with grouping.

Sometimes, we may be given an equation that does not look like a quadratic at first glance. In our next examples we will solve a cubic polynomial equation where the GCF of each term is $x$, and can be factored.  The result is a quadratic equation that we can solve.

### Example

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

In this last video example, we solve a quadratic equation with a leading coefficient of $-1$ using a shortcut method of factoring and the zero product principle.

Quadratic equations can be solved in many ways. You may already be familiar with factoring to solve some quadratic equations. However, not all quadratic equations can be factored. In this topic, you will use square roots to learn another way to solve quadratic equations—and this method will work with all quadratic equations.

# (10.1.2) – Solve a Quadratic Equation by the Square Root Property

One way to solve the quadratic equation $x^{2}=9$ is to subtract 9 from both sides to get one side equal to 0: $x^{2}-9=0$. The expression on the left can be factored, it is a difference of squares: $\left(x+3\right)\left(x–3\right)=0$. Using the zero factor property, you know this means $x+3=0$ or $x–3=0$, so $x=−3$ or 3.

Another property would let you solve that equation more easily is called the square root property.

### The Square Root Property

If $x^{2}=a$, then $x=\sqrt{a}$ or $-\sqrt{a}$.

The property above says that you can take the square root of both sides of an equation, but you have to think about two cases: the positive square root of $a$ and the negative square root of $a$.

A shortcut way to write “$\sqrt{a}$” or “$-\sqrt{a}$” is $\pm \sqrt{a}$. The symbol $\pm$ is often read “positive or negative.” If it is used as an operation (addition or subtraction), it is read “plus or minus.”

### Example

Solve using the Square Root Property. $x^{2}=9$

Notice that there is a difference here in solving $x^{2}=9$ and finding $\sqrt{9}$. For $x^{2}=9$, you are looking for all numbers whose square is 9. For $\sqrt{9}$, you only want the principal (nonnegative) square root. The negative of the principal square root is $-\sqrt{9}$; both would be $\pm \sqrt{9}$. Unless there is a symbol in front of the radical sign, only the nonnegative value is wanted!

In the example above, you can take the square root of both sides easily because there is only one term on each side. In some equations, you may need to do some work to get the equation in this form. You will find that this involves isolating $x^{2}$.

In our first video we will show more examples of using the square root property to solve a quadratic equation.

### Example

Solve. $10x^{2}+5=85$

Sometimes more than just the $x$ is being squared:

### Example

Solve. $\left(x–2\right)^{2}–50=0$

In the next video you will see more examples of using square roots to solve quadratic equations.

# (10.1.3) – Solve a Quadratic Equation by Completing the Square

Not all quadratic equations can be factored or 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.

First, let’s make sure we can recognize a perfect square trinomial and how factor it.

### Example

Factor $9x^{2}–24x+16$.

If this were an equation, we could solve using either the square root property or the zero product property. If you don’t start with a perfect square trinomial, you can complete the square to make what you have into one.

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.

### Steps for Completing The Square

We will use the example ${x}^{2}+4x+1=0$ to illustrate each step.

1. Given a quadratic equation that cannot be factored, and with $a=1$, first add or subtract the constant term to the right side of the equal sign.
${x}^{2}+4x=-1$
2. Multiply the $b$ term by $\frac{1}{2}$ and square it.
$\begin{array}{l}\frac{1}{2}\left(4\right)=2\hfill \\ {2}^{2}=4\hfill \end{array}$
3. Add ${\left(\frac{1}{2}b\right)}^{2}$ to both sides of the equal sign and simplify the right side. We have
$\begin{array}{l}{x}^{2}+4x+4=-1+4\hfill \\ {x}^{2}+4x+4=3\hfill \end{array}$
4. The left side of the equation can now be factored as a perfect square.
$\begin{array}{l}{x}^{2}+4x+4=3\hfill \\ {\left(x+2\right)}^{2}=3\hfill \end{array}$
5. Use the square root property and solve.
$\begin{array}{l}\sqrt{{\left(x+2\right)}^{2}}=\pm \sqrt{3}\hfill \\ x+2=\pm \sqrt{3}\hfill \\ x=-2\pm \sqrt{3}\hfill \end{array}$
6. The solutions are $x=-2+\sqrt{3}$, $x=-2-\sqrt{3}$.

### Example

Solve by completing the square. $x^{2}–12x–4=0$

### Example

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

In the next video you will see more examples of how to use completing the square to solve a quadratic equation.

You may have noticed that because you have to use both square roots, all the examples have two solutions. Here is another example that’s slightly different.

### Example

Solve by completing the square. $x^{2}+16x+17=-47$.

Take a closer look at this problem and you may see something familiar. Instead of completing the square, try adding 47 to both sides in the equation. The equation $x^{2}+16x+17=−47$ becomes $x^{2}+16x+64=0$. Can you factor this equation using grouping? (Think of two numbers whose product is 64 and whose sum is 16).

It can be factored as $(x+8)(x+8)=0$, of course! Knowing how to complete the square is very helpful, but it is not always the only way to solve an equation.

In our last video, we show an example of how to use completing the square to solve a quadratic equation whose solutions are irrational.

# (10.1.4) – The Quadratic Formula

You can solve any quadratic equation by completing the square—rewriting part of the equation as a perfect square trinomial. If you complete the square on the generic equation $ax^{2}+bx+c=0$ and then solve for $x$, you find that $\displaystyle x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. This equation is known as the Quadratic Formula.

We can derive the quadratic formula by completing the square. First, 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:

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

This formula is very helpful for solving quadratic equations that are difficult or impossible to factor, and using it can be faster than completing the square. The Quadratic Formula can be used to solve any quadratic equation of the form $ax^{2}+bx+c=0$.

The form $ax^{2}+bx+c=0$ is called standard form of a quadratic equation. Before solving a quadratic equation using the Quadratic Formula, it’s vital that you be sure the equation is in this form. If you don’t, you might use the wrong values for $a$, $b$, or $c$, and then the formula will give incorrect solutions.

The Quadratic Formula will work with any quadratic equation, but only if the equation is in standard form, $ax^{2}+bx+c=0$. To use it, follow these steps.

• Put the equation in standard form first.
• Identify the coefficients, $a$, $b$, and $c$. Be careful to include negative signs if the $bx$ or $c$ terms are subtracted.
• Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.
• Simplify as much as possible.
• Use the $\pm$ in front of the radical to separate the solution into two values: one in which the square root is added, and one in which it is subtracted.
• Simplify both values to get the possible solutions.

That’s a lot of steps. Let’s try using the Quadratic Formula to solve a relatively simple equation first; then you’ll go back and solve it again using another factoring method.

### Example

Use the quadratic formula to solve the equation $x^{2}+4x=5$.

You can check these solutions by substituting $1$ and $−5$ into the original equation.

 $\begin{array}{r}x=1\\x^{2}+4x=5\\\left(1\right)^{2}+4\left(1\right)=5\\1+4=5\\5=5\end{array}$ $\begin{array}{r}x=-5\\x^{2}+4x=5\,\,\,\,\,\\\left(-5\right)^{2}+4\left(-5\right)=5\,\,\,\,\,\\25-20=5\,\,\,\,\,\\5=5\,\,\,\,\,\end{array}$

You get two true statements, so you know that both solutions work: $x=1$ or $-5$. You’ve solved the equation successfully using the quadratic formula!

Watch this video to see an example of how to use the quadratic formula to solve a quadratic equation that has two real, rational solutions.

Sometimes, it may be easier to solve an equation using conventional factoring methods, like finding number pairs that sum to one number (in this example, 4) and that produce a specific product (in this example $−5$) when multiplied. The power of the Quadratic Formula is that it can be used to solve any quadratic equation, even those where finding number combinations will not work.

In the next video example we show that the quadratic formula is useful when a quadratic equation has two irrational solutions that could not have been obtained by factoring.

Most of the quadratic equations you’ve looked at have two solutions, like the one above. The following example is a little different.

### Example

Use the quadratic formula to solve the equation $x^{2}-2x=6x-16$.

#### Quadratic equations with complex solutions

Next, we will show that some quadratic equations can have complex solutions.  As we simplify with the quadratic formula, we may end up with a negative number under a square root, which, as we know, is not defined for real numbers.

We have seen two outcomes for solutions to quadratic equations, either there was one or two real number solutions. We have also learned that it is possible to take the square root of a negative number by using imaginary numbers. Having this new knowledge allows us to explore one more possible outcome when we solve quadratic equations. Consider this equation:

$2x^2+3x+6=0$

Using the quadratic formula to solve this equation, we first identify $a$, $b$, and $c$.

$a = 2,b = 3,c = 6$

We can place $a$, $b$ and $c$ into the quadratic formula and simplify to get the following result:

$\displaystyle x=-\frac{3}{4}+\frac{\sqrt{-39}}{4}, x=-\frac{3}{4}-\frac{\sqrt{-39}}{4}$

Up to this point, we would have said that $\sqrt{-39}$ is not defined for real numbers and determine that this equation has no solutions.  But, now that we have defined the square root of a negative number, we can also define a solution to this equation as follows.

$\displaystyle x=-\frac{3}{4}+i\frac{\sqrt{39}}{4}, x=-\frac{3}{4}-i\frac{\sqrt{39}}{4}$

In the following example we will work through the process of solving a quadratic equation with complex solutions. Take note that we be simplifying complex numbers – so if you need a review of how to rewrite the square root of a negative number as an imaginary number, now is a good time.

### Example

Use the quadratic formula to solve the equation $x^{2}+2x=-5$.

We can check these solutions in the original equation. Be careful when you expand the squares, and replace $i^{2}$ with $-1$.

 $\begin{array}{r}x=-1+2i\\x^{2}+2x=-5\\\left(-1+2i\right)^{2}+2\left(-1+2i\right)=-5\\1-4i+4i^{2}-2+4i=-5\\1-4i+4\left(-1\right)-2+4i=-5\\1-4-2=-5\\-5=-5\end{array}$ $\begin{array}{r}x=-1-2i\\x^{2}+2x=-5\\\left(-1-2i\right)^{2}+2\left(-1-2i\right)=-5\\1+4i+4i^{2}-2-4i=-5\\1+4i+4\left(-1\right)-2-4i=-5\\1-4-2=-5\\-5=-5\end{array}$

### Example

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

### Example

Use the quadratic formula to solve the equation $x^2+x=-x-3$

# (10.1.5) – Use the discriminant to classify solutions of quadratic equations

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, ${b}^{2}-4ac$, it tells us whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. In turn, we can then determine whether a quadratic function has real or complex roots. 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 repeated rational 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

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

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

We have seen that a quadratic equation may have two real solutions, one real solution, or two complex solutions.

Let’s summarize how the discriminant affects the evaluation of $\sqrt{{{b}^{2}}-4ac}$, and how it helps to determine the solution set.

• If $b^{2}-4ac>0$, then the number underneath the radical will be a positive value. You can always find the square root of a positive, so evaluating the quadratic formula will result in two real solutions (one by adding the positive square root, and one by subtracting it).
• If $b^{2}-4ac=0$, then you will be taking the square root of 0, which is 0. Since adding and subtracting 0 both give the same result, the “$\pm[/late]'' portion of the formula doesn't matter. There will be one real repeated solution. • If [latex]b^{2}-4ac<0$, then the number underneath the radical will be a negative value. Since you cannot find the square root of a negative number using real numbers, there are no real solutions. However, you can use imaginary numbers. You will then have two complex solutions, one by adding the imaginary square root and one by subtracting it.

### Example

Use the discriminant to determine how many and what kind of solutions the quadratic equation $x^{2}-4x+10=0$ has.