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Algebra 1

Course: Algebra 1 > Unit 14

Lesson 3: Solving by taking the square root
Math
Algebra 1
Quadratic functions & equations
Solving by taking the square root
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Solving quadratics by taking square roots

Google Classroom
Learn how to solve quadratic equations like x^2=36 or (x-2)^2=49.

What you should be familiar with before taking this lesson

What you will learn in this lesson

So far you have solved linear equations, which include constant terms—plain numbers—and terms with the variable raised to the first power, x, start superscript, 1, end superscript, equals, x.
You will now learn how to solve quadratic equations, which include terms where the variable is raised to the second power, x, squared.
Here are a few examples of the types of quadratic equations you will learn to solve:
x, squared, equals, 36
left parenthesis, x, minus, 2, right parenthesis, squared, equals, 49
2, x, squared, plus, 3, equals, 131
Now let's get down to business.

Solving x, squared, equals, 36 and similar equations

Suppose we want to solve the equation x, squared, equals, 36. Let's first verbalize what the equation is asking us to find. It is asking us which number, when multiplied by itself, equals 36.
If this question sounds familiar to you, it's because this is the definition of the square root of 36, which is expressed mathematically as square root of, 36, end square root.
Now, this is how the complete solution of the equation looks:
x2=36x2=36Take the square root.x=±36x=±6\begin{aligned}x^2&=36\\\\ \sqrt{x^2}&=\sqrt{36}&&\text{Take the square root.}\\\\ x&=\pm\sqrt{36}\\\\ x&=\pm 6\end{aligned}
Let's review what went on in this solution.

What the plus minus sign means

Note that every positive number has two square roots: a positive square root and a negative square root. For example, both 6 and minus, 6, when squared, equal 36. Therefore, this equation has two solutions.
The plus minus is just an efficient way of representing this concept mathematically. For example, plus minus, 6 means "either 6 or minus, 6".

A note about inverse operations

When we solved linear equations, we isolated the variable by using inverse operations: If the variable had 3 added to it, we subtracted 3 from both sides. If the variable was multiplied by 4, we divided both sides by 4.
The inverse operation of taking the square is taking the square root. However, unlike the other operations, when we take the square root we must remember to take both the positive and the negative square roots.
Now solve a few similar equations on your own.
Problem 1
Solve x, squared, equals, 16.
x, equals, plus minus
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

Problem 2
Solve x, squared, equals, 81.
x, equals, plus minus
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

Problem 3
Solve x, squared, equals, 5.
Choose 1 answer:

Solving left parenthesis, x, minus, 2, right parenthesis, squared, equals, 49 and similar equations

Here is how the solution of the equation left parenthesis, x, minus, 2, right parenthesis, squared, equals, 49 goes:
(x−2)2=49(x−2)2=49Take the square root.x−2=±7x=±7+2Add 2.\begin{aligned}(x-2)^2&=49\\\\ \sqrt{(x-2)^2}&=\sqrt{49}&&\text{Take the square root.}\\\\ x-2&=\pm 7\\\\ x&=\pm 7+2&&\text{Add 2.}\end{aligned}
Therefore, the solutions are x, equals, 9 and x, equals, minus, 5.
Let's review what went on in this solution.

Isolating x

Using the inverse operation of taking the square root, we removed the square sign. This was important in order to isolate x, but we still had to add 2 in the last step in order to really isolate x.

Understanding the solutions

Our work ended with x, equals, plus minus, 7, plus, 2. How should we understand that expression? Remember that plus minus, 7 means "either plus, 7 or minus, 7." Therefore, we should split our answer according to the two cases: either x, equals, 7, plus, 2 or x, equals, minus, 7, plus, 2.
This gives us the two solutions x, equals, 9 and x, equals, minus, 5.
Now solve a few similar equations on your own.
Problem 4
Solve left parenthesis, x, plus, 3, right parenthesis, squared, equals, 25.
Choose 1 answer:

Problem 5
Solve left parenthesis, 2, x, minus, 1, right parenthesis, squared, equals, 9.
Choose 1 answer:

Problem 6
Solve left parenthesis, x, minus, 5, right parenthesis, squared, equals, 7.
Choose 1 answer:

Why we shouldn't expand the parentheses

Let's go back to our example equation, left parenthesis, x, minus, 2, right parenthesis, squared, equals, 49. Suppose we wanted to expand the parentheses there. After all, this is what we do in linear equations, right?
Expanding the parentheses results in the following equation:
x, squared, minus, 4, x, plus, 4, equals, 49
If we wanted to take the square root in this equation, we would have to take the square root of the expression x, squared, minus, 4, x, plus, 4, but it's not clear if square root of, x, squared, minus, 4, x, plus, 4, end square root can be rewritten as a nice expression.
In contrast, taking the square roots of expressions like x, squared or left parenthesis, x, minus, 2, right parenthesis, squared gives us nice expressions like x or left parenthesis, x, minus, 2, right parenthesis.
Therefore, it's actually helpful in quadratic equations to keep things factored, because this allows us to take the square root.

Solving 2, x, squared, plus, 3, equals, 131 and similar equations

Not all quadratic equations are solved by immediately taking the square root. Sometimes we have to isolate the squared term before taking its root.
For example, to solve the equation 2, x, squared, plus, 3, equals, 131 we should first isolate x, squared. We do this exactly as we would isolate the x term in a linear equation.
2x2+3=1312x2=128Subtract 3.x2=64Divide by 2.x2=64Take the square root.x=±8\begin{aligned}2x^2+3&=131\\\\ 2x^2&=128&&\text{Subtract 3.}\\\\ x^2&=64&&\text{Divide by 2.}\\\\ \sqrt{x^2}&=\sqrt{64}&&\text{Take the square root.}\\\\ x&=\pm 8\end{aligned}
Now solve a few similar equations on your own.
Problem 7
Solve 3, x, squared, minus, 7, equals, 5.
Choose 1 answer:

Problem 8
Solve 4, left parenthesis, x, minus, 1, right parenthesis, squared, plus, 2, equals, 38.
Choose 1 answer:

Challenge problem
Solve x, squared, plus, 8, x, plus, 16, equals, 9.
Choose 1 answer:

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