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Study Guides > Intermediate Algebra

The Quadratic Formula

Learning Outcomes

  • 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 all real solutions of a quadratic equation and recognize when there are no real solutions
  • Solve application problems involving quadratic equations
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 [latex]x[/latex], 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 is vital that you be sure the equation is in this form. If you do not, 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.
  1. Put the equation in standard form first.
  2. Identify the coefficients, a, b, and c. Be sure to include negative signs if the bx or c terms are subtracted.
  3. Carefully substitute the values noted in step [latex]2[/latex] into the equation. To avoid needless errors, use parentheses around each number input into the formula.
  4. Simplify as much as possible.
  5. 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.
  6. Simplify both values to get the possible solutions.
That is a lot of steps. Let us try using the Quadratic Formula to solve a relatively simple equation first; then you will 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, [latex]6[/latex] 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]

The solutions are [latex]x=1\,\,\,\text{or}\,\,\,-5[/latex].

  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 have 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, [latex]4[/latex]) 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 have 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 [latex]6[/latex]x from each side and add [latex]16[/latex] 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 [latex]0[/latex] is [latex]0[/latex], and both adding and subtracting [latex]0[/latex] give the same result, there is only one possible value.

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

The answer is [latex]x=4[/latex].

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]

In the next example, we will show that some quadratic equations do not have real 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.

Example

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

Answer: Add [latex]x[/latex] 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] x=\frac{-2\pm \sqrt{-8}}{2}[/latex]

Since the square root of a negative number is not defined for real numbers, there are no real number solutions to this equation.

Applying the Quadratic Formula

Quadratic equations are widely used in science, business, and engineering. Quadratic equations are commonly used in situations where two things are multiplied together and they both depend on the same variable. For example, when working with area, if both dimensions are written in terms of the same variable, you use a quadratic equation. Because the quantity of a product sold often depends on the price, you sometimes use a quadratic equation to represent revenue as a product of the price and the quantity sold. Quadratic equations are also used when gravity is involved, such as the path of a ball or the shape of cables in a suspension bridge. A very common and easy-to-understand application is the height of a ball thrown at the ground off a building. Because gravity will make the ball speed up as it falls, a quadratic equation can be used to estimate its height any time before it hits the ground. Note: The equation is not completely accurate, because friction from the air will slow the ball down a little. For our purposes, this is close enough.

Example

A ball is thrown off a building from [latex]200[/latex] feet above the ground. Its starting velocity (also called initial velocity) is [latex]−10[/latex] feet per second. The negative value means it is heading toward the ground. The equation [latex]h=-16t^{2}-10t+200[/latex] can be used to model the height of the ball after t seconds. About how long does it take for the ball to hit the ground?

Answer: When the ball hits the ground, the height is [latex]0[/latex]. Substitute [latex]0[/latex] for h.

[latex]\begin{array}{c}h=-16t^{2}-10t+200\\0=-16t^{2}-10t+200\\-16t^{2}-10t+200=0\end{array}[/latex]

This equation is difficult to solve by factoring or by completing the square, so solve it by applying the Quadratic Formula, [latex] x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex]. In this case, the variable is t rather than x. [latex]a=−16,b=−10[/latex], and [latex]c=200[/latex].

[latex] t=\frac{-(-10)\pm \sqrt{{{(-10)}^{2}}-4(-16)(200)}}{2(-16)}[/latex]

Simplify. Be very careful with the signs.

[latex] \begin{array}{l}t=\frac{10\pm \sqrt{100+12800}}{-32}\\\,\,=\frac{10\pm \sqrt{12900}}{-32}\end{array}[/latex]

Use a calculator to find both roots.

t is approximately [latex]−3.86[/latex] or [latex]3.24[/latex].

Consider the roots logically. One solution, [latex]−3.86[/latex], cannot be the time because it is a negative number. The other solution, [latex]3.24[/latex] seconds, must be when the ball hits the ground. The ball hits the ground approximately [latex]3.24[/latex] seconds after being thrown.

In the next video, we show another example of how the quadratic equation can be used to find the time it takes for an object in free fall to hit the ground. https://youtu.be/RcVeuJhcuL0 The area problem below does not look like it includes a Quadratic Formula of any type, and the problem seems to be something you have solved many times before by simply multiplying. But in order to solve it, you will need to use a quadratic equation.

Example

Bob made a quilt that is [latex]4[/latex] ft [latex]\times[/latex] [latex]5[/latex] ft. He has [latex]10[/latex] sq. ft. of fabric he can use to add a border around the quilt. How wide should he make the border to use all the fabric? (The border must be the same width on all four sides.)

Answer: Sketch the problem. Since you do not know the width of the border, you will let the variable [latex]x[/latex] represent the width. In the diagram. The original quilt is indicated by the red rectangle. The border is the area between the red and blue lines. A blue rectangle. Within the blue rectangle are a pair of vertical parallel lines and a pair of horizontal parallel lines that create a smaller red rectangle. The lengths of this red rectangle are 4 feet and 5 feet. The line segments between the boundaries of the red rectangle and the bigger blue rectangle are all labeled x. Since each side of the original [latex]4[/latex] by [latex]5[/latex] quilt has the border of width x added, the length of the quilt with the border will be [latex]5+2x[/latex], and the width will be [latex]4+2x[/latex]. (Both dimensions are written in terms of the same variable, and you will multiply them to get an area! This is where you might start to think that a quadratic equation might be used to solve this problem.) A blue rectangle with one side a height of 4+2x and another side a length of 5+2x. Within the blue rectangle are a pair of vertical parallel lines and a pair of horizontal parallel lines that create a smaller red rectangle. The height of this red rectangle is 4 feet and the length is 5 feet. The line segments between the boundaries of the red rectangle and the bigger blue rectangle are all labeled x. You are only interested in the area of the border strips. Write an expression for the area of the border.

Area of border = Area of the blue rectangle minus the area of the red rectangle

Area of border[latex]=\left(4+2x\right)\left(5+2x\right)–\left(4\right)\left(5\right)[/latex]

There are [latex]10[/latex] sq ft of fabric for the border, so set the area of border to be [latex]10[/latex].

[latex]10=\left(4+2x\right)\left(5+2x\right)–20[/latex]

Multiply [latex]\left(4+2x\right)\left(5+2x\right)[/latex].

[latex]10=20+8x+10x+4x^{2}–20[/latex]

Simplify.

[latex]10=18x+4x^{2}[/latex]

Subtract [latex]10[/latex] from both sides so that you have a quadratic equation in standard form and can apply the Quadratic Formula to find the roots of the equation.

[latex]\begin{array}{c}0=18x+4x^{2}-10\\\\\text{or}\\\\4x^{2}+18x-10=0\\\\2\left(2x^{2}+9x-5\right)=0\end{array}[/latex]

Factor out the greatest common factor, [latex]2[/latex], so that you can work with the simpler equivalent equation, [latex]2x^{2}+9x–5=0[/latex].

[latex]\begin{array}{r}2\left(2x^{2}+9x-5\right)=0\\\\\frac{2\left(2x^{2}+9x-5\right)}{2}=\frac{0}{2}\\\\2x^{2}+9x-5=0\end{array}[/latex]

Use the Quadratic Formula. In this case, [latex]a=2,b=9[/latex], and [latex]c=−5[/latex].

[latex]\begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\frac{-9\pm \sqrt{{{9}^{2}}-4(2)(-5)}}{2(2)}\end{array}[/latex]

Simplify.

[latex] x=\frac{-9\pm \sqrt{121}}{4}=\frac{-9\pm 11}{4}[/latex]

Find the solutions, making sure that the [latex]\pm[/latex] is evaluated for both values.

[latex]\begin{array}{c}x=\frac{-9+11}{4}=\frac{2}{4}=\frac{1}{2}=0.5\\\\\text{or}\\\\x=\frac{-9-11}{4}=\frac{-20}{4}=-5\end{array}[/latex]

Ignore the solution [latex]x=−5[/latex], since the width could not be negative. The width of the border should be [latex]0.5[/latex] ft.

Our last video gives another example of using the quadratic formula for a geometry problem involving the border around a quilt. https://youtu.be/Zxe-SdwutxA

Summary

Quadratic equations can appear in different applications. The Quadratic Formula is a useful way to solve these equations or any other quadratic equation! The Quadratic Formula, [latex] x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex], is found by completing the square of the quadratic equation [latex] [/latex]. When you simplify using the quadratic formula and your result is a negative number under a square root, there are no real number solutions to the equation.

Licenses & Attributions

CC licensed content, Original

  • 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

  • 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) for Lumen Learning. License: CC BY: Attribution.