Solving Radical Equations

To solve a radical equation, perform inverse operations in the usual way. But take note: = | a|, and thus expressions such as must be solved as absolute value expressions for more on solving equations containing absolute values. It is not necessary to solve ()2 as an absolute value expression.


Example: Solve for x: (x + 5)2 = 18.

= .
| x + 5| = 3.
x + 5 = 3 or x + 5 = - 3.
x = 3 - 5 or x = - 3 - 5.

Since we cannot take the square root of a negative number, there are often numbers which appear to be solutions but do not actually make the equation true. For example, we get two solutions when we solve = - 2x:

()2 = (- 2x)2.
6x + 10 = 4x2.
4x2 - 6x - 10 = 0.
2(2x2 - 3x - 5) = 0.
2(x + 1)(2x - 5) = 0.
x = - 1 or .

We can plug -1 in for x in the original equation to check that it makes the equation true:

= - 2(- 1)?
= 2? Yes.
True.

However, when we plug in for x:

= - 2()?
= - 5? No.
False.

is an extraneous solution, and the only solution to = - 2x is x = - 1.

As demonstrated by this example, we must check all "solutions" and eliminate false solutions.