Solving Equations That Have Radical Terms

Part 1

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

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. $(\sqrt{x})^2 = 5^2$ $x = 25$ 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. 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 $a = 1, b = -11, c = 25$ $x = \dfrac{11 \pm \sqrt{121 - 4(1)(25)}}{2 (1)}$ $x = \dfrac{11 \pm \sqrt{21}}{2}$

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

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.