### 13.2 - Radical equations

Before reading this section you may want to review the following topics:
A **radical equation** is one in which the unknown, call it *x*, is inside a radical.
To solve a radical equation follow these steps:
- Isolate the term containing the radical on one side of the equation,
say on the left-hand-side. (By isolate we mean get it by itself.)

- If the radical is a square root then square both sides of the equation.
(In general, if the radical is an
*n*^{th} root then take the
*n*^{th} power of both sides.)
On the left-hand-side use the fact that
to get rid of the radical. On the right-hand-side distribute.

- The unknown
*x* is no longer inside a radical.
Now you can finish solving for *x*
by using the basic procedures for solving equations.

**Note:**
- If the equation contains more than one radical term, you will have to
perform the above procedure several times.

- One of the steps, squaring both sides of the equation, increases the degree of
the unknown, and this often leads to extraneous solutions. Therefore it is
**very important to check your solutions**.

**Example: ** Solve the radical equation .

**Solution:** The first step is to isolate the radical term:
Then square both sides:
6 *x* + 4 = 4 *x*^{ 2}.

The result is that this is no longer a radical equation; it is a quadratic equation.
Divide both sides by 2 to simplify it and then collect all terms on one side of the
equation to put it into standard form:
2 *x*^{ 2} − 3 *x* − 2 = 0.

Click here to see how the left-hand-side of
this quadratic equation can be factored. The result is this:
(*x* − 2) (2 *x* + 1) = 0.

The purpose of factoring is to put the equation into the form
*a* · *b* = 0.
We can now replace it with two new equations. Each new equation comes from setting
one of the factors to zero:
*x* − 2 = 0

2 *x* + 1 = 0

Their solutions are *x* = 2 and *x* = −½.
**We must check these solutions.** Substituting *x* = 2 into the original
equation and simplifying causes the equation to read 0 = 0, so this solution checks out.
But substituting *x* = −½ back into the original equation
causes it to read 2 = 0, so this solution doesn’t check out and must be rejected.
Thus this radical equation has the single solution *x* = 2.

**Example: ** Solve the radical equation .

**Solution:** The first step is to isolate one of the radical terms
(it doesn’t matter which one):
Then square both sides:
On the left-hand-side squaring got rid of the radical.
On the right-hand-side we will have to distribute.
After distributing and collecting like terms we get this equation:
This is still a radical equation because although one radical is gone,
the other one still remains. So we repeat the entire process.
Isolate the remaining radical term (by canceling *x* terms
and dividing by 32):
Then square both sides:
*x* − 32 = 49

Note that this is no longer a radical equation; both radicals are now gone.
Solving this equation yields the solution *x* = 81. **We must now check the solution.**
Substituting *x* = 81 back into the original equation and simplifying gives
the equation 16 = 16, so this solution checks out.

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