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Solving Equations That Have Radical Terms

Part 1

In this video, we are going to solve equations with a radical expression.

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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.

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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

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.