Radical Equations

We need to be extra careful when solving radical equations. It’s necessary to verify the final “solutions” against the original equation because there’s a high likelihood of obtaining extraneous answers.

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 just \(x=4\).