Key Questions

A radical equation is any equation that involves any type of radical (square root, cube root, fourth root, et cetera).

Explanation:

Examples of radicals are: $\sqrt{46}$ and ³sqrt(81). A radical equation is an equation that comprises one or more radicals.

$x - 6 = \sqrt{x}$

This can be solved easily by:

${x}^{2} - 12 x + 36 = x$

${x}^{2} - 13 x + 36 = 0$

$\left(x - 9\right) \left(x - 4\right) = 0$

$x = 9 , x = 4$

but wait! we have to be careful whenever we square both sides of an equation.

Re-substitute the two answers into the original equation.

$9 - 6 = \sqrt{9}$
$3 = 3$

That's fine, one more.

$4 - 6 = \sqrt{4}$
$- 2 \ne 2$

Our extraneous answer is $x = 4$

To find the extraneous answer of any equation, just find the solutions and then re-substitute the solutions into the original equation. The answer that results in an illogical equation is the extraneous answer.

• Use the facts (1) the cube of the cube root of an expression is equal to the expression and (2) cubing both sides of an equation yields an equivalent equation.
That is: (1) ${\left(\sqrt[3]{a}\right)}^{3} = a$ and (2) $a = b$ if and only if ${a}^{3} = {b}^{3}$.

(Note that point 2, above does NOT apply to squares. Squaring may introduce additional solutions. E.g. the only solution to $x = 3$ is the obvious one. But ${x}^{2} = 9$ has two solutions.)

Here's an example of you question:

Solve $\sqrt[3]{2 x + 3} = 5$.

This equation is equivalent to (has the same solutions as): ${\left(\sqrt[3]{2 x + 3}\right)}^{3} = {\left(5\right)}^{3}$ .

Which simplifies to: $2 x + 3 = 125$
And this is true exactly when $2 x = 122$
Which has only one solution: $61$.

(Or, if you insist, it has solution $x = 61$.)

• A radical equation is solved by simply squaring or cubing both sides of the equation. For example:

√(2x+9)=5

√(2x+9)^2 = ${5}^{2}$

Then, continue to isolate the variable:

$2 x + 9 = 25$

$2 x + 9 - 9 = 25 - 9$

$\frac{2 x}{2} = \frac{16}{2}$

$x = 8$