We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

TEXT
Unlock Solution Steps
Sign in to
Symbolab
Get full access to all Solution Steps for any math problem
By continuing, you agree to our Terms of Use
and have read our Privacy Policy

Study Guides > College Algebra: Co-requisite Course

Quadratic Equations

Learning Objectives

  • Recognize a quadratic equation
    • Use the zero product principle to solve quadratic equations that can be factored
    • Identify solutions to quadratic equations on a graph
  • Square Roots and Completing the Square
    • Use the square root property to solve a quadratic equation
    • Complete the square to solve a quadratic equation
  • The Quadratic Formula
    • Write a quadratic equation in standard form and identify the values of a, b, and c in a standard form quadratic equation.
    • Use the Quadratic Formula to find solutions of a quadratic equation, (rational, irrational and complex)
An equation containing a second-degree polynomial is called a quadratic equation. For example, equations such as [latex]2{x}^{2}+3x - 1=0[/latex] and [latex]{x}^{2}-4=0[/latex] 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 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 [latex]a\cdot b=0[/latex], then [latex]a=0[/latex] or [latex]b=0[/latex], 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, [latex]a{x}^{2}+bx+c=0[/latex], where a, b, and c are real numbers, and [latex]a\ne 0[/latex]. The equation [latex]{x}^{2}+x - 6=0[/latex] 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
[latex]\text{If }a\cdot b=0,\text{ then }a=0\text{ or }b=0[/latex],
where a and b are real numbers or algebraic expressions. A quadratic equation is an equation containing a second-degree polynomial; for example
[latex]a{x}^{2}+bx+c=0[/latex]
where a, b, and c are real numbers, and if [latex]a\ne 0[/latex], it is in standard form.

Solving Quadratics with a Leading Coefficient of 1

In the quadratic equation [latex]{x}^{2}+x - 6=0[/latex], the leading coefficient, or the coefficient of [latex]{x}^{2}[/latex], 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 [latex]\left(x+k\right)\text{ or }\left(x-k\right)[/latex], 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 [latex]-2[/latex], the factors are [latex]\left(x+1\right)\left(x - 2\right)[/latex].
  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: [latex]{x}^{2}+x - 6=0[/latex].

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

[latex]\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}[/latex]
The last pair, [latex]3\cdot \left(-2\right)[/latex] sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.
[latex]\left(x - 2\right)\left(x+3\right)=0[/latex]
To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.
[latex]\begin{array}{l}\left(x - 2\right)\left(x+3\right)\hfill&=0\hfill \\ \left(x - 2\right)\hfill&=0\hfill \\ x\hfill&=2\hfill \\ \left(x+3\right)\hfill&=0\hfill \\ x\hfill&=-3\hfill \end{array}[/latex]
The two solutions are [latex]x=2[/latex] and [latex]x=-3[/latex]. If we graph the function [latex]f(x)={x}^{2}+x - 6[/latex], we will get the parabola in the figure below. The solutions to the equation[latex]{x}^{2}+x - 6=0[/latex] are the x-intercepts of the function [latex]f(x)={x}^{2}+x - 6[/latex]. Recall that x-intercepts are where the outputs, or y values are zero, therefore the points (-3,0) and (2,0) represent the places where the parabola crosses the x axis. Quadratic functions will be discussed in further detail in the next section. 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.

In the following video we provide more examples of factoring to solve quadratic equations. https://youtu.be/bi7i_RuIGl0   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: [latex]{x}^{2}-9=0[/latex].

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.

[latex]\begin{array}{l}{x}^{2}-9=0\hfill \\ \left(x - 3\right)\left(x+3\right)=0\hfill \\ \hfill \\ \left(x - 3\right)=0\hfill \\ x=3\hfill \\ \hfill \\ \left(x+3\right)=0\hfill \\ x=-3\hfill \end{array}[/latex]
The solutions are [latex]x=3[/latex] and [latex]x=-3[/latex].

 Solving Quadratics with a Leading Coefficient of [latex]\ne1[/latex]

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, [latex]a{x}^{2}+bx+c=0[/latex], multiply [latex]a\cdot c[/latex].
  2. Find two numbers whose product equals [latex]ac[/latex] and whose sum equals [latex]b[/latex].
  3. Rewrite the equation replacing the [latex]bx[/latex] 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: [latex]4{x}^{2}+15x+9=0[/latex].

Answer: First, multiply [latex]ac:4\left(9\right)=36[/latex]. Then list the factors of [latex]36[/latex].

[latex]\begin{array}{l}1\cdot 36\hfill \\ 2\cdot 18\hfill \\ 3\cdot 12\hfill \\ 4\cdot 9\hfill \\ 6\cdot 6\hfill \end{array}[/latex]
The only pair of factors that sums to [latex]15[/latex] is [latex]3+12[/latex]. Rewrite the equation replacing the b term, [latex]15x[/latex], with two terms using 3 and 12 as coefficients of x. Factor the first two terms, and then factor the last two terms.
[latex]\begin{array}{l}4{x}^{2}+3x+12x+9=0\hfill \\ x\left(4x+3\right)+3\left(4x+3\right)=0\hfill \\ \left(4x+3\right)\left(x+3\right)=0\hfill \end{array}[/latex]
Solve using the zero-product property.
[latex]\begin{array}{l}\left(4x+3\right)\left(x+3\right)=0\hfill \\ \hfill \\ \left(4x+3\right)=0\hfill \\ x=-\frac{3}{4}\hfill \\ \hfill \\ \left(x+3\right)=0\hfill \\ x=-3\hfill \end{array}[/latex]
The solutions are [latex]x=-\frac{3}{4}[/latex], [latex]x=-3[/latex]. Coordinate plane with the x-axis ranging from negative 6 to 2 with every other tick mark labeled and the y-axis ranging from negative 6 to 2 with each tick mark numbered. The equation: four x squared plus fifteen x plus nine is graphed with its x-intercepts: (-3/4,0) and (-3,0) plotted as well.

The following video contains another example of solving a quadratic equation using factoring with grouping. https://youtu.be/04zEXaOiO4U 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: [latex]-3{x}^{3}-5{x}^{2}-2x=0[/latex].

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 [latex]-x[/latex] from all of the terms and then proceed with grouping.

[latex]\begin{array}{l}-3{x}^{3}-5{x}^{2}-2x=0\hfill \\ -x\left(3{x}^{2}+5x+2\right)=0\hfill \end{array}[/latex]
Use grouping on the expression in parentheses.
[latex]\begin{array}{l}-x\left(3{x}^{2}+3x+2x+2\right)=\hfill&0\hfill \\ -x\left[3x\left(x+1\right)+2\left(x+1\right)\right]=\hfill&0\hfill \\ -x\left(3x+2\right)\left(x+1\right)=\hfill&0\hfill \end{array}[/latex]
Now, we use the zero-product property. Notice that we have three factors.
[latex]\begin{array}{l}-x\hfill&=0\hfill \\ x\hfill&=0\hfill \\ 3x+2\hfill&=0\hfill \\ x\hfill&=-\frac{2}{3}\hfill \\ x+1\hfill&=0\hfill \\ x\hfill&=-1\hfill \end{array}[/latex]
The solutions are [latex]x=0[/latex], [latex]x=-\frac{2}{3}[/latex], and [latex]x=-1[/latex].

  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. https://youtu.be/nZYfgHygXis 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.

Solve a Quadratic Equation by the Square Root Property

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

The Square Root Property

If [latex]x^{2}=a[/latex], then [latex] x=\sqrt{a}[/latex] or [latex] -\sqrt{a}[/latex]. 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 “[latex] \sqrt{a}[/latex]” or “[latex] -\sqrt{a}[/latex]” is [latex] \pm \sqrt{a}[/latex]. The symbol [latex]\pm[/latex] 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. [latex]x^{2}=9[/latex]

Answer: Since one side is simply [latex]x^{2}[/latex], you can take the square root of both sides to get x on one side. Don’t forget to use both positive and negative square roots!

[latex]\begin{array}{l}x^{2}=9\\\,\,\,x=\pm\sqrt{9}\\\,\,\,x=\pm3\end{array}[/latex]

Answer

[latex]x=\pm3[/latex] (that is, [latex]x=3[/latex] or [latex]-3[/latex])

Notice that there is a difference here in solving [latex]x^{2}=9[/latex] and finding [latex] \sqrt{9}[/latex]. For [latex]x^{2}=9[/latex], you are looking for all numbers whose square is 9. For [latex] \sqrt{9}[/latex], you only want the principal (nonnegative) square root. The negative of the principal square root is [latex] -\sqrt{9}[/latex]; both would be [latex] \pm \sqrt{9}[/latex]. 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 [latex]x^{2}[/latex]. In our first video we will show more examples of using the square root property to solve a quadratic equation. https://youtu.be/Fj-BP7uaWrI

Example

Solve. [latex]10x^{2}+5=85[/latex]

Answer: If you try taking the square root of both sides of the original equation, you will have [latex] \sqrt{10{{x}^{2}}+5}[/latex] on the left, and you can’t simplify that. Subtract 5 from both sides to get the [latex]x^{2}[/latex] term by itself.

[latex]10x^{2}+5=85[/latex]

You could now take the square root of both sides, but you would have [latex] \sqrt{10}[/latex] as a coefficient, and you would need to divide by that coefficient. Dividing by 10 before you take the square root will be a little easier.

[latex]10x^{2}=80[/latex]

Now you have only [latex]x^{2}[/latex] on the left, so you can use the Square Root Property easily. Be sure to simplify the radical if possible.

[latex] \begin{array}{l}{{x}^{2}}=8\\\,\,\,x=\pm \sqrt{8}\\\,\,\,\,\,\,=\pm \sqrt{(4)(2)}\\\,\,\,\,\,\,=\pm \sqrt{4}\sqrt{2}\\\,\,\,\,\,\,=\pm 2\sqrt{2}\end{array}[/latex]

Answer

[latex-display] x=\pm 2\sqrt{2}[/latex-display]

Sometimes more than just the x is being squared:

Example

Solve. [latex]\left(x–2\right)^{2}–50=0[/latex]

Answer: Again, taking the square root of both sides at this stage will leave something you can’t work with on the left. Start by adding 50 to both sides.

[latex]\left(x-2\right)^{2}-50=0[/latex]

Because [latex]\left(x–2\right)^{2}[/latex] is a squared quantity, you can take the square root of both sides.

[latex]\begin{array}{r}\left(x-2\right)^{2}=50\,\,\,\,\,\,\,\,\,\,\\x-2=\pm\sqrt{50}\end{array}[/latex]

To isolate x on the left, you need to add 2 to both sides. Be sure to simplify the radical if possible.

[latex] \begin{array}{l}x=2\pm \sqrt{50}\\\,\,\,\,=2\pm \sqrt{(25)(2)}\\\,\,\,\,=2\pm \sqrt{25}\sqrt{2}\\\,\,\,\,=2\pm 5\sqrt{2}\end{array}[/latex]

Answer

[latex-display] x=2\pm 5\sqrt{2}[/latex-display]

In the next video you will see more examples of using square roots to solve quadratic equations. https://youtu.be/4H5qZ_-8YM4

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 [latex]9x^{2}–24x+16[/latex].

Answer: First notice that the [latex]x^{2}[/latex] term and the constant term are both perfect squares. [latex-display]\begin{array}{l}9x^{2}=\left(3x\right)^{2}\\\,\,\,16=4^{2}\end{array}[/latex-display] Then notice that the middle term (ignoring the sign) is twice the product of the square roots of the other terms. [latex-display]24x=2\left(3x\right)\left(4\right)[/latex-display] A trinomial in the form [latex]r^{2}-2rs+s^{2}[/latex] can be factored as [latex](r–s)^{2}[/latex]. In this case, the middle term is subtracted, so subtract r and s and square it to get [latex](r–s)^{2}[/latex]. [latex-display]\begin{array}{c}\,\,\,r=3x\\s=4\\9x^{2}-24x+16=\left(3x-4\right)^{2}\end{array}[/latex-display]

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 [latex]{x}^{2}+4x+1=0[/latex] to illustrate each step.
  1. Given a quadratic equation that cannot be factored, and with [latex]a=1[/latex], first add or subtract the constant term to the right side of the equal sign.
    [latex]{x}^{2}+4x=-1[/latex]
  2. Multiply the b term by [latex]\frac{1}{2}[/latex] and square it.
    [latex]\begin{array}{l}\frac{1}{2}\left(4\right)=2\hfill \\ {2}^{2}=4\hfill \end{array}[/latex]
  3. Add [latex]{\left(\frac{1}{2}b\right)}^{2}[/latex] to both sides of the equal sign and simplify the right side. We have
    [latex]\begin{array}{l}{x}^{2}+4x+4=-1+4\hfill \\ {x}^{2}+4x+4=3\hfill \end{array}[/latex]
  4. The left side of the equation can now be factored as a perfect square.
    [latex]\begin{array}{l}{x}^{2}+4x+4=3\hfill \\ {\left(x+2\right)}^{2}=3\hfill \end{array}[/latex]
  5. Use the square root property and solve.
    [latex]\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}[/latex]
  6. The solutions are [latex]x=-2+\sqrt{3}[/latex], [latex]x=-2-\sqrt{3}[/latex].

Example

Solve by completing the square. [latex]x^{2}–12x–4=0[/latex]

Answer: Since you cannot factor the trinomial on the left side, you will use completing the square to solve the equation. First, move the constant term to the right side of the equal sign.Identify b.

[latex]\begin{array}{r}x^{2}-12x=4\,\,\,\,\,\,\,\,\\b=-12\end{array}[/latex]

Then, take [latex]\frac{1}{2}[/latex] of the b term and square it. Add [latex] {{\left( \frac{b}{2}\right)}^{2}}[/latex] to complete the square, so [latex] {{\left( \frac{b}{2} \right)}^{2}}={{\left( \frac{-12}{2} \right)}^{2}}={{\left( -6 \right)}^{2}}=36[/latex]. Add the value to both sides of the equation and simplify.

[latex]\begin{array}{l}x^{2}-12x+36=4+36\\x^{2}-12x+36=40\end{array}[/latex]

Rewrite the left side as a squared binomial.

[latex]\left(x-6\right)^{2}=40[/latex]

Use the Square Root Property. Remember to include both the positive and negative square root, or you’ll miss one of the solutions.

[latex] x-6=\pm\sqrt{40}[/latex]

Solve for x by adding 6 to both sides. Simplify as needed.

[latex] \begin{array}{l}x=6\pm \sqrt{40}\\\,\,\,\,=6\pm \sqrt{4}\sqrt{10}\\\,\,\,\,=6\pm 2\sqrt{10}\end{array}[/latex]

Answer

[latex-display] x=6\pm 2\sqrt{10}[/latex-display]

Example

Solve by completing the square: [latex]{x}^{2}-3x - 5=0[/latex].

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

[latex]{x}^{2}-3x=5[/latex]
Identify b.[latex]b=-3[/latex] Then, take [latex]\frac{1}{2}[/latex] of the b term and square it.
[latex]\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}[/latex]
Add the result to both sides of the equal sign.
[latex]\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}[/latex]
Factor the left side as a perfect square and simplify the right side.
[latex]{\left(x-\frac{3}{2}\right)}^{2}=\frac{29}{4}[/latex]
Use the square root property and solve.
[latex]\begin{array}{l}\sqrt{{\left(x-\frac{3}{2}\right)}^{2}}\hfill&=\pm \sqrt{\frac{29}{4}}\hfill \\ \left(x-\frac{3}{2}\right)\hfill&=\pm \frac{\sqrt{29}}{2}\hfill \\ x\hfill&=\frac{3}{2}\pm \frac{\sqrt{29}}{2}\hfill \end{array}[/latex]
The solutions are [latex]x=\frac{3}{2}+\frac{\sqrt{29}}{2}[/latex], [latex]x=\frac{3}{2}-\frac{\sqrt{29}}{2}[/latex].

In the next video you will see more examples of how to use completing the square to solve a quadratic equation. https://youtu.be/PsbYUySRjFo 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. [latex]x^{2}+16x+17=-47[/latex].

Answer: Rewrite the equation so the left side has the form [latex]x^{2}+bx[/latex]. Identify b.

[latex]\begin{array}{c}x^{2}+16x=-64\\b=16\end{array}[/latex]

Add [latex] {{\left( \frac{b}{2} \right)}^{2}}[/latex], which is [latex] {{\left( \frac{16}{2} \right)}^{2}}={{8}^{2}}=64[/latex], to both sides.

[latex]\begin{array}{l}x^{2}+16x+64=-64+64\\x^{2}+16x+64=0\end{array}[/latex]

Write the left side as a squared binomial.

[latex]\left(x+8\right)^{2}=0[/latex]

Take the square roots of both sides. Normally both positive and negative square roots are needed, but 0 is neither positive nor negative. 0 has only one root.

[latex]x+8=0[/latex]

Solve for x.

[latex]x=-8[/latex]

Answer

[latex-display]x=-8[/latex-display]

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 [latex]x^{2}+16x+17=−47[/latex] becomes [latex]x^{2}+16x+64=0[/latex]. 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 [latex](x+8)(x+8)=0[/latex], 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. https://youtu.be/IjCjbtrPWHM

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 [latex]ax^{2}+bx+c=0[/latex] and then solve for x, you find that [latex]x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex]. 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 [latex]-1[/latex] and obtain a positive a. Given [latex]a{x}^{2}+bx+c=0[/latex], [latex]a\ne 0[/latex], we will complete the square as follows:
  1. First, move the constant term to the right side of the equal sign:
    [latex]a{x}^{2}+bx=-c[/latex]
  2. As we want the leading coefficient to equal 1, divide through by a:
    [latex]{x}^{2}+\frac{b}{a}x=-\frac{c}{a}[/latex]
  3. Then, find [latex]\frac{1}{2}[/latex] of the middle term, and add [latex]{\left(\frac{1}{2}\frac{b}{a}\right)}^{2}=\frac{{b}^{2}}{4{a}^{2}}[/latex] to both sides of the equal sign:
    [latex]{x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}}=\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}[/latex]
  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:
    [latex]{\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}[/latex]
  5. Now, use the square root property, which gives
    [latex]\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}[/latex]
  6. Finally, add [latex]-\frac{b}{2a}[/latex] to both sides of the equation and combine the terms on the right side. Thus,
    [latex]x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex]
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 [latex]ax^{2}+bx+c=0[/latex]. The form [latex]ax^{2}+bx+c=0[/latex] 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.

Solving a Quadratic Equation using the Quadratic Formula

The Quadratic Formula will work with any quadratic equation, but only if the equation is in standard form, [latex]ax^{2}+bx+c=0[/latex]. 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 [latex]\pm[/latex] 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 [latex]x^{2}+4x=5[/latex].

Answer: First write the equation in standard form.

[latex]\begin{array}{r}x^{2}+4x=5\,\,\,\\x^{2}+4x-5=0\,\,\,\\\\a=1,b=4,c=-5\end{array}[/latex]

Note that the subtraction sign means the constant c is negative.

[latex] \begin{array}{r}{{x}^{2}}\,\,\,+\,\,\,4x\,\,\,-\,\,\,5\,\,\,=\,\,\,0\\\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\a{{x}^{2}}\,\,\,+\,\,\,bx\,\,\,+\,\,\,c\,\,\,=\,\,\,0\end{array}[/latex]

Substitute the values into the Quadratic Formula. [latex] x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\[/latex]

[latex] \begin{array}{l}\\x=\frac{-4\pm \sqrt{{{(4)}^{2}}-4(1)(-5)}}{2(1)}\end{array}[/latex]

Simplify, being careful to get the signs correct.

[latex]x=\frac{-4\pm\sqrt{16+20}}{2}[/latex]

Simplify some more.

[latex] x=\frac{-4\pm \sqrt{36}}{2}[/latex]

Simplify the radical: [latex] \sqrt{36}=6[/latex].

[latex] x=\frac{-4\pm 6}{2}[/latex]

Separate and simplify to find the solutions to the quadratic equation. Note that in one, 6 is added and in the other, 6 is subtracted.

[latex]\begin{array}{c}x=\frac{-4+6}{2}=\frac{2}{2}=1\\\\\text{or}\\\\x=\frac{-4-6}{2}=\frac{-10}{2}=-5\end{array}[/latex]

Answer

[latex-display]x=1\,\,\,\text{or}\,\,\,-5[/latex-display]

You can check these solutions by substituting [latex]1[/latex] and [latex]−5[/latex] into the original equation.
[latex]\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}[/latex] [latex]\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}[/latex]
You get two true statements, so you know that both solutions work: [latex]x=1[/latex] or [latex]-5[/latex]. 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. https://youtu.be/xtwO-n8lRPw 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 [latex]−5[/latex]) 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. https://youtu.be/tF0muV86dr0 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 [latex]x^{2}-2x=6x-16[/latex].

Answer: Subtract 6x from each side and add 16 to both sides to put the equation in standard form.

[latex]\begin{array}{l}x^{2}-2x=6x-16\\x^{2}-2x-6x+16=0\\x^{2}-8x+16=0\end{array}[/latex]

Identify the coefficients a, b, and c. [latex]x^{2}=1x^{2}[/latex], so [latex]a=1[/latex]. Since [latex]8x[/latex] is subtracted, b is negative. [latex]a=1,b=-8,c=16[/latex]

[latex] \begin{array}{r}{{x}^{2}}\,\,\,-\,\,\,8x\,\,\,+\,\,\,16\,\,\,=\,\,\,0\\\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\a{{x}^{2}}\,\,\,+\,\,\,bx\,\,\,+\,\,\,\,c\,\,\,\,=\,\,\,0\end{array}[/latex]

Substitute the values into the Quadratic Formula.

[latex]\begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\frac{-(-8)\pm \sqrt{{{(-8)}^{2}}-4(1)(16)}}{2(1)}\end{array}[/latex]

Simplify.

[latex] x=\frac{8\pm \sqrt{64-64}}{2}[/latex]

Since the square root of 0 is 0, and both adding and subtracting 0 give the same result, there is only one possible value.

[latex] x=\frac{8\pm \sqrt{0}}{2}=\frac{8}{2}=4[/latex]

Answer

[latex-display]x=4[/latex-display]

Again, check using the original equation.

[latex]\begin{array}{r}x^{2}-2x=6x-16\,\,\,\,\,\\\left(4\right)^{2}-2\left(4\right)=6\left(4\right)-16\\16-8=24-16\,\,\,\,\,\,\\8=8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

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:

[latex]2x^2+3x+6=0[/latex]

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

[latex]a = 2,b = 3,c = 6[/latex]

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

[latex]x=-\frac{3}{4}+\frac{\sqrt{-39}}{4}, x=-\frac{3}{4}-\frac{\sqrt{-39}}{4}[/latex]

Up to this point, we would have said that [latex]\sqrt{-39}[/latex] 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.

[latex]x=-\frac{3}{4}+i\frac{\sqrt{39}}{4}, x=-\frac{3}{4}-i\frac{\sqrt{39}}{4}[/latex]

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 [latex]x^{2}+2x=-5[/latex].

Answer: First write the equation in standard form. [latex-display]\begin{array}{r}x^{2}+2x=-5\\x^{2}+2x+5=0\,\,\,\,\,\\\\a=1,b=2,c=5\,\,\,\end{array}[/latex-display] [latex-display] \begin{array}{r}{{x}^{2}}\,\,\,+\,\,\,2x\,\,\,+\,\,\,5\,\,\,=\,\,\,0\\\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\a{{x}^{2}}\,\,\,+\,\,\,bx\,\,\,+\,\,\,c\,\,\,=\,\,\,0\end{array}[/latex-display] Substitute the values into the Quadratic Formula. [latex-display] x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\x=\frac{-2\pm \sqrt{{{(2)}^{2}}-4(1)(5)}}{2(1)}[/latex-display] Simplify, being careful to get the signs correct. [latex-display] x=\frac{-2\pm \sqrt{4-20}}{2}[/latex-display] Simplify some more. [latex-display] x=\frac{-2\pm \sqrt{-16}}{2}[/latex-display] Simplify the radical, but notice that the number under the radical symbol is negative! The square root of [latex]−16[/latex] is imaginary. [latex] \sqrt{-16}=4i[/latex]. [latex-display] x=\frac{-2\pm 4i}{2}[/latex-display] Separate and simplify to find the solutions to the quadratic equation. [latex-display]\begin{array}{c}x=\frac{-2+4i}{2}=\frac{-1+2i}{1}\cdot \frac{2}{2}=-1+2i\\\\\text{or}\\\\x=\frac{-2-4i}{2}=\frac{-1-2i}{1}\cdot \frac{2}{2}=-1-2i\end{array}[/latex-display]

Answer

[latex-display]x=-1+2i[/latex] or [latex]-1-2i[/latex-display]

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

Example

Use the quadratic formula to solve [latex]{x}^{2}+x+2=0[/latex].

Answer: First, we identify the coefficients: [latex]a=1,b=1[/latex], and [latex]c=2[/latex]. Substitute these values into the quadratic formula. [latex]\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}[/latex] Now we can separate the expression [latex]\frac{-1\pm i\sqrt{7}}{2}[/latex] into two solutions: [latex-display]-\frac{1}{2}+\frac{ i\sqrt{7}}{2}[/latex-display] [latex-display]-\frac{1}{2}-\frac{ i\sqrt{7}}{2}[/latex-display]   The solutions to the equation are [latex]x=\frac{-1+i\sqrt{7}}{2}[/latex] and [latex]x=\frac{-1-i\sqrt{7}}{2}[/latex] or [latex]x=\frac{-1}{2}+\frac{i\sqrt{7}}{2}[/latex] and [latex]x=\frac{-1}{2}-\frac{i\sqrt{7}}{2}[/latex].

 

Example

Use the quadratic formula to solve the equation [latex]x^2+x=-x-3[/latex]

Answer: Add x to both sides, and add 3 to both sides to get the quadratic equation in standard form.

[latex]\begin{array}{l}x^{2}+x=-x-3\\x^{2}+2x+3=0\end{array}[/latex]

Identify a, b, c.

[latex]a=1, b=2, c=3[/latex]

Substitute values for a, b, c into the quadratic formula.

[latex]\begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\frac{-2\pm \sqrt{{{(2)}^{2}}-4(1)(3)}}{2(1)}\end{array}[/latex]

Simplify

[latex]\displaystyle x=\frac{-2\pm \sqrt{-8}}{2}[/latex]

Rewrite the radical of a negative number in terms of the imaginary unit [latex]i[/latex]

[latex]\displaystyle x=\frac{-2\pm i\sqrt{8}}{2}[/latex]

Simplify the radical

[latex]\displaystyle x=\frac{-2\pm 2i\sqrt{2}}{2}[/latex]

Reduce the fraction

[latex]x=-1 \pm i\sqrt{2}[/latex]

 Answer

[latex-display]x=-1 \pm i\sqrt{2}[/latex-display]

Try It

[ohm_question]31110[/ohm_question]
 

Licenses & Attributions

CC licensed content, Original

  • Revision and Adaptation. Provided by: Lumen Learning License: CC BY: Attribution.
  • Quadratic Formula Application - Time for an Object to Hit the Ground. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Quadratic Formula Application - Determine the Width of a Border. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.

CC licensed content, Shared previously

  • College Algebra. Provided by: OpenStax Authored by: Abramson, Jay, et al.. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution. License terms: Download for free at : http://cnx.org/contents/[email protected]:1/Preface.
  • Ex: Solve a Quadratic Equation Using Factor By Grouping. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • Ex: Factor and Solve Quadratic Equation - Trinomial a = -1. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • Unit 12: Factoring, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education Located at: https://www.nroc.org/. License: CC BY: Attribution.
  • Ex 1: Solving Quadratic Equations Using Square Roots. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • Ex 2: Solving Quadratic Equations Using Square Roots. Authored by: James Sousa (Mathispower4u.com) . License: Public Domain: No Known Copyright.
  • Ex 1: Completing the Square - Real Rational Solutions. Authored by: James Sousa (Mathispower4u.com) . License: Public Domain: No Known Copyright.
  • Ex 2: Completing the Square - Real Irrational Solutions. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • Ex2: Quadratic Formula - Two Real Irrational Solutions. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Ex: Quadratic Formula - Two Real Rational Solutions. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • Question ID#31110. Authored by: Wallace,Tyler, mb Sousa,James. License: CC BY: Attribution.