An equation in which the unknown appears in a radicand is called a radical equation.

For instance: $\sqrt{{{x^2} – 5}} = x + 1$ and $\sqrt {3t + 7} + \sqrt {t + 2} = 1$ are radical equations.

To solve a radical equation, begin by isolating the most complicated radical expression on one side of the equation, and then eliminate the radical by raising both sides of the equation to a power equal to the index of the radical. You may have to repeat this technique in order to eliminate all radicals. When the equation is radical-free, simplify and solve it. Since extraneous roots may be introduced when both sides of an equation are raised to even powers, all roots must be checked by substitution in the original equation whenever a radical with an even index is involved.

Example:

Solve $\sqrt{{x – 1}} – 2 = 0$.

Solution:

To isolate the radical, we add $2$ to both sides of the equation
$\sqrt{{x – 1}} = 2$

Now we raise both sides to the power $3$ and obtain
${\left( {\sqrt{{x – 1}}} \right)^3} = {2^3}$     or   $x – 1 = 8$

It follows that $x = 9$. Since we did not raise both sides of the equation to an even power, it is not necessary to check our solution, but it is good practice to do so anyway. Substituting $x = 9$ in the left side of the original equation, we obtain
$\sqrt{{x – 1}} – 2 = \sqrt{{9 – 1}} – 2 = 2 – 2 = 0$

Hence, the solution $x = 9$ is correct.

Example:

Solve $\sqrt {3t + 7} + \sqrt {t + 2} = 1$.

Solution:

We add $– \sqrt {t + 2}$to both sides of the equation to isolate $\sqrt {3t + 7}$ on the left side. Thus
$\sqrt {3t + 7} = 1 – \sqrt {t + 2}$

Now we square both sides of the equation to obtain
${\left( {\sqrt {3t + 7} } \right)^2} = {\left( {1 – \sqrt {t + 2} } \right)^2}$
$3t + 7 = 1 – 2\sqrt {t + 2} + \left( {t + 2} \right)$

The equation still contains a radical, so we simplify and isolate this radical
$2t + 4 = – 2\sqrt {t + 2}$
$– \left( {t + 2} \right) = \sqrt {t + 2}$

Again, we square both sides, so we get
${\left[ { – \left( {t + 2} \right)} \right]^2} = {\left( {\sqrt {t + 2} } \right)^2}$
${t^2} + 4t + 4 = t + 2$
${t^2} + 3t + 2 = 0$
$\left( {t + 2} \right)\left( {t + 1} \right) = 0$
$t = – 2$ or $t = – 1$

To Check: $t = – 2$

$\sqrt {3t + 7} + \sqrt {t + 2} = \sqrt {3\left( { – 2} \right) + 7} + \sqrt { – 2 + 2} = \sqrt 1 + \sqrt 0 = 1$
Hence, $t = – 2$ is a solution.

To Check: $t = – 1$

$\sqrt {3t + 7} + \sqrt {t + 2} = \sqrt {3\left( { – 1} \right) + 7} + \sqrt { – 1 + 2} = \sqrt 4 + \sqrt 1 \ne 1$

Therefore, $t = – 1$ is an extraneous root that was introduced by squaring both sides of the equation. The only solution $t = – 2$