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2 Tutorials that teach Solving a Quadratic Equation using the Quadratic Formula
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Solving a Quadratic Equation using the Quadratic Formula

Solving a Quadratic Equation using the Quadratic Formula

Author: Colleen Atakpu
Description:

In this lesson, students will learn more in depth about how to use the quadratic formula to solve quadratic equations.

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[MUSIC PLAYING] Let's look at our objectives for today. We'll start by reviewing quadratic equations. We'll then review the quadratic formula, which is used to solve quadratic equations. And finally, we'll do some examples solving quadratic equations with the quadratic formula.

Let's start by reviewing quadratic equations. A quadratic equation is an equation that can be written in the form ax squared plus bx plus c equals 0, where a, b, and c are real numbers. Factoring, or variable isolation, may be used to solve some quadratic equations, but not all. However, the quadratic formula can be used to find solutions to all quadratic equations, even when factoring or variable isolation is difficult or impossible. Therefore, sometimes it is necessary to use the quadratic formula to find solutions to a quadratic equation.

Now let's review the quadratic formula. The quadratic formula says that the solutions to a quadratic equation, x, are equal to negative b plus or minus the square root of b squared minus 4ac, all over 2a. The values for a, b, and c in the quadratic formula come from the values of a, b, and c in the quadratic equation. The plus-minus symbol here indicates that a quadratic equation may have two solutions. And when working with the quadratic formula, we often have to simplify square roots using the product property, or by recognizing perfect squares.

Now let's do some examples. We want to solve the quadratic equation x squared plus 7x minus 4 equals 0. We can solve using the quadratic formula. From the equation, we see that there is no written number in front of x squared, meaning that there is an implied coefficient of 1. Therefore, a equals 1. We also see that b equals 7 and c equals negative 4.

Substituting these values into the formula gives us negative 7 plus or minus the square root of 7 squared minus 4 times 1 times negative 4 over 2 times 1. We can simplify the numerator and the denominator separately. Simplifying the denominator is simple. 2 times 1 is 2, so we now have this expression.

We then simplify the numerator, which is more complicated, because it involves the plus-minus symbol, square roots, and other operations. We start underneath the square root. 7 squared is 49, and 4 times 1 times negative 4 is negative 16. So we have 49 minus negative 16, which is the same as 49 plus 16, which is 65. The square root of 65 cannot be further simplified, so we leave it as written.

We now have x equals negative 7 plus or minus the square root of 65 over 2. We can separate the solution into its two parts by separating the plus and the minus symbols. First solution is x equals negative 7 plus the square root of 65 over 2. And the second solution is negative 7 minus the square root of 65 over 2.

We can further simplify both of these fractions into two separate fractions each. So the first solution would be x equals negative 7 over 2 plus the square root of 65 over 2. And the second solution, x equals negative 7 over 2 minus the square root of 65 over 2.

Here's our second example. We want to solve the equation 2x squared minus 8x minus 3 equals 0. From the equation, we see that a equals 2, b equals negative 8, and c equals negative 3. Notice that b is already a negative number, so when we substitute it into the formula, we have a negative negative 8, which is positive 8.

We'll start simplifying in the denominator. 2 times 2 gives us 4. We then simplify in the numerator, starting with our square root. Negative 8 squared is a positive 64. And 4 times 2 times negative 3 is negative 24. So we have 64 minus negative 24, or 64 plus 24, which is 88.

We can simplify the square root of 88 as the square root of 4 times the square root of 22 using the product property of radicals. The square root of 4 is 2, so we have 2 square root of 22. So our solution is x equals 8 plus or minus 2 square root of 22 over 4. We can separate this into the two solutions. 8 plus 2 squared of 22 over 4, and 8 minus 2 square root of 22 over 4.

We can also separate each of these solutions into separate fractions. 8 over 4 plus 2 square root of 22 over 4, and 8 over 4 minus 2 square root of 22 over 4. In this form, we can see that we can simplify these fractions further. 8 over 4 is 2, and 2 squared of 22 over 4 can be simplified by dividing the 2 and 4 both by 2, which gives a squared of 22 over 2. So our fully simplified solution is x equals 2 plus square root of 22 over 2, and x equals 2 minus square root of 22 over 2.

Let's go over our important points from today. Make sure you get these in your notes so you can refer to them later. Factoring, or variable isolation, may be used to solve some, but not all, quadratic equations. However, the quadratic formula can be used to find solutions to all quadratic equations, even when factoring or variable isolation is difficult or impossible. Before using the quadratic formula, the equation must be equal to 0 in order to determine the correct values of a, b, and c to use in the formula.

So I hope that these important points and examples helped you understand a little bit more about solving a quadratic equation using the quadratic formula. Keep using your notes and keep on practicing and soon you'll be a pro. Thanks for watching.

Notes on "Solving a Quadratic Equation using the Quadratic Formula"

00:00 – 00:33 Introduction

00:34 – 01:15 Quadratic Equations

01:16 – 01:56 The Quadratic Formula

01:57 – 06:09 Examples Solving Quadratic Equations using Quadratic Formula

06:10 – 06:58 Important to Remember (Recap)

TERMS TO KNOW
  • KEY FORMULA

    x = [-b±sqrt(b^2-4ac)]/2a

  • KEY FORMULA

    sqrt(ab) = sqrt(a)sqrt(b)