Part 1

In this video, we are going to solve equations with a radical expression. After you finish this lesson, view all of our Algebra 1 lessons and practice problems.

In \sqrt{x}=7, first square both sides of the equation
\sqrt{x}^2=7^2
x=49

If there are other values in the equation, then first isolate the \sqrt{x} before squaring both sides of the equation.

Video-Lesson Transcript

Let’s go over how to solve equations with a radical expression.

We have

\sqrt{x} = 7

In order to solve for x, we have to get rid of the radical sign.

The inverse of a square root is squaring.

So we’ll square both sides

(\sqrt{x})^2 = 7^2
x = 49

If we have something like this:

\sqrt{x} + 3 = 8

we want to get the radical by itself first.

Let’s subtract 3 on both sides

\sqrt{x} + 3 - 3 = 8 - 3
\sqrt{x} = 5

And the last step is to square both sides.

(\sqrt{x})^2 = 5^2
x = 25
Solving Equations That Have Radical Terms

Let’s look at a more complicated one

3 \sqrt{x} - 4 = 7

Again, we want to have the radical by itself. Let’s add 4 on both sides.

3 \sqrt{x} - 4 + 4 = 7 + 4
3 \sqrt{x} = 11
\dfrac{3 \sqrt{x}}{3} = \dfrac{11}{3}
\sqrt{x} = \dfrac{11}{3}
(\sqrt{x})^2 = (\dfrac{11}{3})^2
x = \dfrac{121}{9}

Let’s have this:

\sqrt{x - 8} = 3

Here, the whole term is under the radical, so it is already isolated.

So, we’ll start by squaring both sides.

(\sqrt{x - 8})^2 = 3^2
x - 8 = 9

Now, let’s solve for x.

x - 8 + 8 = 9 + 8
x = 17

 

Part 2

In the second part, we are going to solve equations with a radical expression that are more complex.

For example:
\sqrt{x}=(x-5)

Square both sides of the equation
\sqrt{x}^2=(x-5)^2

Use FOIL to complete the squaring if necessary
x=(x-5)(x-5)
x=x^2-5x-5x+25

Combine like terms
x=x^2-10x+25

Since it’s a quadratic equation, use the quadratic equation
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Substitute the variables
x=\frac{11\pm\sqrt{(-11)^2-4(1)(25)}}{2(1)}
x=\frac{11\pm\sqrt{121-4(1)(25)}}{2(1)}
x=\frac{11\pm\sqrt{21}}{2}

So we have:
x=\frac{11+\sqrt{21}}{2} and x=\frac{11-\sqrt{21}}{2}

In the end, x is approximately 7.8 or 3.2

If it was -7.8 instead of 7.8, then it cannot be a negative answer since the square root of a negative number is imaginary. In this case, the final answer would just be 3.2.

Solving Equations That Have Radical Terms 2

Examples of Solving Equations That Have Radical Terms

Example 1

\sqrt{2x-2}=x-1

First, square both sides of the equation to eliminate the radical symbol
(\sqrt{2x-2})^2=(x-1)^2
Then, solve the equation that comes out after the squaring process
2x-2=(x-1)(x-1)
2x-2=x^2-2x+1

Get all the terms on one side so the other side be zero.
2x-2=x^2-2x+1
2x-2+2=x^2-2x+1+2
2x=x^2-2x+3
2x-2x=x^2-2x+3-2x
0=x^2-4x+3
Now, we can factor out the quadratic function
0=(x-1)(x-3)
Let’s solve for x
x-1=0
x=1
and
x-3=0
x=3
The answers are
(1, 3)

Example 2

\sqrt{5x+11}-1=x
First, add -1 from both sides
\sqrt{5x+11}-1+1=x+1
\sqrt{5x+11}=x+1
Now, we square both sides of the equation to eliminate the radical symbol
(\sqrt{5x+11})^2=(x+1)^2
Then, solve the equation that comes out after the squaring process
5x+11=(x+1)(x+1)
5x+11=x^2+2x+1

Get all the terms on one side so the other side be zero.
5x+11-11=x^2+2x+1-11
5x=x^2+2x-10
5x-5x=x^2+2x-10-5x
0=x^2-3x-10
Now, we can factor out the quadratic function
0=(x+2)(x-5)
Let’s solve for x
x+2=0
x=-2
and
x-5=0
x=5
Remember: if one of the answers come out as a negative, that’s a rejected answer.
Therefore, our final answer is 5

Video-Lesson Transcript

This is how to solve equations with a radical expression part 2.

\sqrt{x} = x - 5

It’s a little bit different because we have x on both sides.

Here, we want to get the radical by itself first before we combine the x together.

In order to get rid of the radical, let’s square both sides.

(\sqrt{x})^2 = (x - 5)^2
x = (x - 5) (x - 5)
x = x^2 - 5x - 5x + 25
x = x^2 - 10x + 25

Now, we have a quadratic equation.

We want to get all the terms on one side so the other side be zero.

x - x = x^2 - 10x - x + 25
0 = x^2 - 11x + 25

This can not be factored so let’s just use the quadratic function to solve it.

Our formula is

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Here we have

a = 1, b = -11, c = 25
x = \dfrac{11 \pm \sqrt{121 - 4(1)(25)}}{2 (1)}
x = \dfrac{11 \pm \sqrt{21}}{2}

Our answers are

{\dfrac{11 + \sqrt{21}}{2}, \dfrac{11 - \sqrt{21}}{2}}

What is important here is that we always want to make sure that we won’t end up with a negative radical.

Let’s find out what is the exact value of this is

\sqrt{21} = 4.6
{\dfrac{11 + 4.6}{2} = 7.8}
{\dfrac{11 - 4.6}{2} = 3.2}

If one of the answers come out as a negative, that’s a rejected answer.

The only answer we accept is a positive number. Because you cannot find the square root of a negative number.

So, you always have to check and make sure that you don’t have a negative number under the radical. After you finish this lesson, view all of our Algebra 1 lessons and practice problems.