Solving Quadratic Equations with the Quadratic Formula
The Quadratic Formula: Given a quadratic equation in the following form:
...where a, b, and c are the numerical coefficients of the terms of the quadratic, the value of the variable x is given by the following equation:
The nice thing about the Quadratic Formula is that the Quadratic Formula always works. There are some quadratics (most of them, actually) that we can't solve by factoring. But the Quadratic Formula will always spit out an answer, whether or not the quadratic expression was factorable.
Let's try that first problem from the previous page again, but this time we'll use the Quadratic Formula instead of the laborious process of completing the square:
Use the Quadratic Formula to solve x2 – 4x – 8 = 0
The Quadratic Formula requires that I have the quadratic expression on one side of the "equals" sign, with "zero" on the other side. They've given me the equation already in that form. Also, the Formula is stated in terms of the numerical coefficients of the terms of the quadratic expression. Looking at the coefficients in this equation, I see that a = 1, b = –4, and c = –8. I'll plug these numbers into the Formula, and simplify. (I should get the same answer as I previously have.)
This is the same answer as I got before, which confirms that the Quadratic Formula works as intended. Once again, my final answer is:
The nice thing about the Quadratic Formula (as compared to completing the square) is that we're just plugging into a formula. There are no "steps" to remember, and thus there are fewer opportunities for mistakes. That being said:
Take care not to omit the "±" sign in front of the radical.
Don't draw the fraction line as being only under the square root, because it's under the initial "–b" part, too.
Don't forget that the denominator of the Formula is "2a", not just "2". That is, when the leading term is something like "5x2", you will need to remember to put the "a = 5" value in the denominator.
Use parentheses around the coefficients when you're first plugging them into the Formula, especially when any of those coefficients is negative, so you don't lose any "minus" signs.
Solve 4x2 + 3x – 2 = 0 using the Quadratic Formula.
First, I'll read off the values of the coefficients that I'll be plugging into the Formula:
a = 4
b = 3
c = – 2
Now all I have to do is plug these values into the Formula, and simplify to get my answer:
Absolutely nothing will simplify here, so I'm done. My answer is:
You should definitely memorize the Quadratic Formula. I don't care if your teacher says she's going to give it to you on the next test; memorize it anyway, because you'll be needing it later. It's not that long, and there's even a song to help you remember it, set to the tune of "Pop Goes the Weasel":
X is equal to negative B
Plus or minus the square root
Of B-squared minus four A C
All over two A
(The above song isn't original to me. I learned it elsewhere.)
When using the Formula, take the time to be careful because, as long as you do your work neatly, the Quadratic Formula will give you the right answer every time.
I have a lesson on the Quadratic Formula, which provides worked examples and shows the connection between the discriminant (the "b2 – 4ac" part inside the square root), the number and type of solutions of the quadratic equation, and the graph of the related parabola. If you're wanting more help with the Formula, then please study the lesson at the above hyperlink.
You can use the Mathway widget below to practice solving quadratic equations by using the Quadratic Formula. Try the entered exercise, or type in your own exercise. Then click the button and select "Solve using the Quadratic Formula" to compare your answer to Mathway's. (Or skip the widget and continue on the next page.)
(Click "Tap to view steps" to be taken directly to the Mathway site for a paid upgrade.)
Somebody (possibly in seventh-century India) was solving a lot of quadratic equations by completing the square. At some point, he (and, yes, it would have been a guy back then) noticed that he was always doing the exact same steps in the exact same order for every equation.
The great power of algebra is that it provides us with the ability to deal with abstractions, such as formulas that always work. This can relieve us from the burden and messiness of having to muck about with the numbers every single time we do the exact same thing. Using this power with respect to solving quadratics by completing the square, he made a formula out of what he'd been doing; namely, the Quadratic Formula, which states:
Solve by Factoring Lessons
Several previous lessons explain the techniques used to factor expressions. This lesson focuses on an imporatant application of those techniques - solving equations.
Why solve by factoring?
The most fundamental tools for solving equations are addition, subtraction, multiplication, and division. These methods work well for equations like x + 2 = 10 - 2x and 2(x - 4) = 0.
But what about equations where the variable carries an exponent, like x2 + 3x = 8x - 6? This is where factoring comes in. We will use this equation in the first example.
The Solve by Factoring process will require four major steps:
- Move all terms to one side of the equation, usually the left, using addition or subtraction.
- Factor the equation completely.
- Set each factor equal to zero, and solve.
- List each solution from Step 3 as a solution to the original equation.
x2 + 3x = 8x - 6
The first step is to move all terms to the left using addition and subtraction. First, we will subtract 8x from each side.
x2 + 3x - 8x = 8x - 8x - 6
x 2 - 5x = -6
Now, we will add 6 to each side.
x2 - 5x + 6 = -6 + 6
x 2 - 5x + 6 = 0
With all terms on the left side, we proceed to Step 2.
We identify the left as a trinomial, and factor it accordingly:
(x - 2)(x - 3) = 0
We now have two factors, (x - 2) and (x - 3).
We now set each factor equal to zero. The result is two subproblems:
x - 2 = 0
x - 3 = 0
Solving the first subproblem, x - 2 = 0, gives x = 2. Solving the second subproblem, x - 3 = 0, gives x = 3.
The final step is to combine the two previous solutions, x = 2 and x = 3, into one solution for the original problem.
x2 + 3x = 8x - 6
x = 2, 3
Solve by Factoring: Why does it work?
Examine the equation below:
ab = 0
If you let a = 3, then logivally b must equal 0. Similarly, if you let b = 10, then a must equal 0.
Now try letting a be some other non-zero number. You should observe that as long as a does not equal 0, b must be equal to zero.
To state the observation more generally, "If ab = 0, then either a = 0 or b = 0." This is an important property of zero which we exploit when solving by factoring.
When the example was factored into (x - 2)(x - 3) = 0, this property was applied to determine that either (x - 2) must equal zero, or (x - 3) must equal zero. Therefore, we were able to create two equations and determine two solutions from this observation.
A Second Example
5x3 = 45x
Move all terms to the left side of the equation. We do this by subtracting 45x from each side.
5x3 - 45x = 45x - 45x
5x 3 - 45x = 0.
The next step is to factor the left side completely. We first note that the two terms on the left have a greatest common factor of 5x.
5x(x2 - 9) = 0
Now, (x2 - 9) can be factored as a difference between two squares.
5x(x + 3)(x - 3) = 0
We are left with three factors: 5x, (x + 3), and (x - 3). As explained in the "Why does it work?" section, at least one of the three factors must be equal to zero.
Create three subproblems by setting each factor equal to zero.
1. 5x = 0
2. x + 3 = 0
3. x - 3 = 0
Solving the first equation gives x = 0. Solving the second equation gives x = -3. And solving the third equation gives x= 3.
The final solution is formed from the solutions to the three subproblems.
x = -3, 0, 3
3x4 - 288x2 - 1200 = 0
Steps 1 and 2
All three terms are already on the left side of the equation, so we may begin factoring. First, we factor out a greatest common factor of 3.
3(x4 - 96x2 - 400) = 0
Next, we factor a trinomial.
3(x2 + 4)(x2 - 100) = 0
Finally, we factor the binomial (x2 - 100) as a difference between two squares.
3(x2 + 4)(x + 10)(x - 10) = 0
We proceed by setting each of the four factors equal to zero, resulting in four new equations.
1. 3 = 0
2. x2 + 4 = 0
3. x + 10 = 0
4. x - 10 = 0
The first equation is invalid, and does not yield a solution. The second equation cannot be solved using basic methods. (x2 + 4 = 0 actually has two imaginary number solutions, but we will save Imaginary Numbers for another lesson!) Equation 3 has a solution of x = -10, and Equation 4 has a solution of x = 10.
We now include all the solutions we found in a single solution to the original problem:
x = -10, 10
This may be abbreviated as
x = ±10
Solve By Factoring Resources
Equation Calculator - Solve By Factoring
Practice Problems / Worksheet
Next Lesson: Quadratic Equations
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