# Making Sense of Radical Equations

Just as students learn to solve linear equations and quadratic equations, they will also need to learn
how to solve radical equations. To solve equations involving radicals, we must use what is known as the power rule, which states that if x, y and n are real numbers and x = y, then xn = yn. This is important because to solve radical equations we have to remove the radical. In order to remove the radical, we must raise the radical to the index of the radical. For example, if we have 3√(x+3), we raise 3√(x+3) to the 3rd power to remove the radical. To remove a radical involving a square root, we raise it to the 2nd power and so on.

Sometimes when we raise both sides of an equation to the same power, we will get equations that are not equivalent. The reason for this is that their solution sets are different.

For example, if x = 4, the solution set is {4}. Now if we raise both sides to the 4th power, we get x4 = 256, with a solution set of {4, -4}. Notice the solution sets are different. The first equation is not satisfied by x = -4, as it is in the second equation. Therefore, we must always check each solution in the original equation to make sure it satisfies the equation. Any solution that doesn't satisfy the equation is discarded. Those discarded solutions are also known as extraneous solutions.

√(x - 3) = 8, √(2x + 5) = √(x + 2), 3√(8x) = 9, 3√(3x + 4) = 3√(x2).

The first step in solving a radical equation is the same as for solving a linear equation. We want to isolate the radical expression to one side of the equation. Next, we raise both sides of the equation to the power that equals the index of the radical. Then we solve the equation for the variable and check all solutions to make sure they satisfy the original equation. Eliminate those solutions which don't satisfy the equation.

Example: Solve √(2x + 7) = 7.

First step is to isolate the radical to one side. The problem is set up with it already isolated. Next step is to square both sides to remove the radical.

[√(2x + 7)]2 = (7)2
2x + 7 = 49

Now solve for x.
2x = 42
x = 21.

Check to make sure 21 satisfies the original equation.
√(2x + 7) = 7
√(2∙21 + 7) = 7
√(42 + 7) = 7
√49 = 7
7 = 7. Therefore 21 is the solution to the equation.

Example: Solve √(4x - 11) + 4 = x.

√(4x - 11) + 4 = x
√(4x - 11) + 4 - 4 = x - 4
√(4x - 11) = x - 4

Next, square both sides of the equation to remove the radical.
[√(4x - 11)]2 = (x - 4)2
4x - 11 = (x - 4)2
Now we solve the equation for x.
4x - 11 = x2 - 8x + 16 (by the FOIL method (x - 4)(x - 4) )
-11 = x2 - 12x + 16
0 = x2 - 12x + 27

Factor to get
0 = (x - 3)(x - 9) (-9 ∙ (-3) = 27 and -9 + (-3) = -12)
0 = (x - 3) or 0 = (x - 9). Therefore x = 3, 9

Check both answers to make sure they satisfy the original equation.
For x = 3, √(4 ∙ 3 - 11) + 4 = 3 For x = 9, √(4 ∙ 9 - 11) + 4 = 9
√1 + 4 = 3 √25 + 4 = 9
5 ≠ 3, 9 = 9

Therefore 9 is a solution to the equation, but 3 is not a solution.

Example: Solve 3√(x3 + 8) = x + 2.
The radical is already isolated on the left side of the equation, so now we have to raise both sides to the 3rd power to remove the radical.

[3√(x3 + 8)]3 = (x + 2)3
x3 + 8 = (x + 2)(x + 2)(x + 2)
x3 + 8 = (x2 + 4x + 4)(x + 2)
x3 + 8 = x3 + 6x2 + 12x + 8
8 = 6x2 + 12x + 8
0 = 6x2 + 12x
0 = 6x(x + 2)
0 = 6x or 0 = x + 2
x = 0, -2

Check solutions in original equation.
For x = 0, 3√(03 + 8) = 0 + 2 For x = -2 3√( ( -2)3 + 8) = -2 + 2
2 = 2, 0 = 0. Therefore, 0 and -2 are both solutions to the equation.

Example: Solve √(2x + 1) + √x = 5.

First we have to isolate a radical on one side. Subtract √x from both sides of the equation.
√(2x + 1) = 5 - √x

Now square both sides to remove the radical from the left side of the equation.
[√(2x + 1)]2 = (5 - √x)2
2x + 1 = 25 - 10√x + x (Note that (5 - √x)2 = (5 - √x)(5 - √x) = 25 - 5√x - 5√x + √(x)2 )

Since we still have a radical in the equation, we need to isolate the radical again.
2x + 1 - 25 = - 10√x + x
2x - 24 = -10√x + x
2x - x - 24 = -10√x
x - 24 = -10√x

Now square both sides of the equation to remove the radical.
(x - 24)2 = (-10√x)2
x2 - 48x + 576 = 100x (Notice that (x - 24)2 = (x - 24)(x - 24) and (-10√x)2 = (-10)(-10)(√x)(√x) )
x2 - 148x + 576 = 0
(x - 144)(x - 4) = 0 (Note that you can factor 576 into 2 ∙ 288 = 2 ∙ 2 ∙ 144 = 4 ∙ 144 and -4 + (-144) = -148).

x - 144 = 0 or x - 4 = 0
x = 4, 144

When substituting the values back into the original equation, 144 does not work, but 4 does. Therefore, 4 is a solution to the equation.

This guide to radical equations with examples solved in step by step fashion should ease any confusion on this topic.