7. Solving Radical Equations and Inequalities - 6. Inverses and Radical Functions and Relations

Solving a equation with an expression raised to a rational exponent usually involves three main steps.

Isolate the expression raised to a rational exponent on one side of the equation.
Raise each side of the equation to a power equal to the reciprocal of the rational exponent.
Solve the resulting equation. Remember to check your results!

Now we can analyze the given equation.

(x - 5)^{\frac{1}{3}} - 4 = - 2

First, let's isolate the expression raised to a rational exponent,

(x - 5)^{\frac{1}{3}},

on one side of the equation.

(x - 5)^{\frac{1}{3}} - 4 = - 2

AddEqn

$LHS + 4 = RHS + 4$

(x - 5)^{\frac{1}{3}} = 2

In order to remove the

\frac{1}{3}

power, or cube root, we will raise each side of the equation to the power of

3 .

(x - 5)^{\frac{1}{3}} = 2

RaiseEqn

$LHS^{3} = RHS^{3}$

[(x - 5)^{\frac{1}{3}}]^{3} = 2^{3}

▼

Solve for

x

PowPow

$(a^{m})^{n} = a^{m \cdot n}$

(x - 5)^{1} = 2^{3}

CalcPow

Calculate power

x - 5 = 8

AddEqn

$LHS + 5 = RHS + 5$

x = 13

Next, we will check the solution by substituting

13

for

x

into the original equation. If the substitution produces a true statement, we know that our answer is correct. If it does not, then it is an extraneous solution.