We need to be extra careful when solving radical equations. The "solutions" that we get in the end must be verified with the original equation. The reason is that the possibility of getting extraneous answers is high.

**Example 1: **Solve the radical equation \(5\sqrt {x + 7} = 25\).

**STEP 1**: Divide both sides by 5.

$$\eqalign{

5\sqrt {x + 7} &= 25 \cr

{{5\sqrt {x + 7} } \over 5} &= {{25} \over 5} \cr

\sqrt {x + 7} &= 5 \cr} $$

**STEP 2**: Take the square of both sides.

$$\eqalign{

{\left( {\sqrt {x + 7} } \right)^2} &= {5^2} \cr

x + 7 &= 25 \cr} $$

**STEP 3**: Subtract both sides by 7.

$$\eqalign{

x + 7 &= 25 \cr

x +7- 7 &= 25 - 7 \cr

x &= 18 \cr} $$

Check if indeed \(x = 18\) is a solution.

$$\eqalign{

5\sqrt {x + 7} &= 25 \cr

5\sqrt {\left( {18} \right) + 7} &= 25 \cr

5\sqrt {25} &= 25 \cr

5\left( 5 \right) &= 25 \cr

25 &= 25 \cr} $$

Since it yields a true statement, we say that \(x = 18\) is a solution.

**Example 2: **Solve the radical equation \(\sqrt {33 - 2x} = x + 1\).

**STEP 1:** Square both sides. The radical symbol on the left is gone while we will be squaring the binomial on the right side. We are left with a quadratic equation to solve. Try the factoring method first before using the Quadratic Formula as it is a mess.

$$\eqalign{

{\left( {\sqrt {33 - 2x} } \right)^2} &= {\left( {x + 1} \right)^2} \cr

33 - 2x &= {x^2} + 2x + 1 \cr} $$

**STEP 2:** Let's move everything to the right side. Then, we simplify.

$$\eqalign{

0 &= {x^2} + 2x + 2x + 1 - 33 \cr

0 &= {x^2} + 4x - 32 \cr} $$

**STEP 3:** We factor the trinomial on the right-hand side.

$$0 = \left( {x - 4} \right)\left( {x + 8} \right)$$

**STEP 4:** Set each factor equal to zero then solve.

$$\eqalign{

x - 4 &= 0 \cr

x &= 4 \cr} $$

or

$$\eqalign{

x + 8 &= 0 \cr

x &= - 8 \cr} $$

**STEP 5:** Check each "solution" if it is a valid answer by substituting it back to the original radical equation.

For \(x=4\),

$$\eqalign{

\sqrt {33 - 2x} &= x + 1 \cr

\sqrt {33 - 2\left( 4 \right)} &= \left( 4 \right) + 1 \cr

\sqrt {33 - 8} &= 5 \cr

\sqrt {25} &= 5 \cr

5 &= 5 \cr} $$

Yes, that means \(x=4\) is a valid solution!

For \(x=-8\),

$$\eqalign{

\sqrt {33 - 2x} &= x + 1 \cr

\sqrt {33 - 2\left( { - 8} \right)} &= - 8 + 1 \cr

\sqrt {33 + 16} &= - 7 \cr

\sqrt {49} &= - 7 \cr

7 &\ne - 7 \cr} $$

This tells us that \(x=-8\) is NOT a solution.

Therefore, the final answer is simply \(x=4\).