# How do you solve  (x-2)^(3/4)=8?

Mar 25, 2016

${\left(x - 2\right)}^{\frac{3}{4}} = 8$

There is an easy solution which is:

${\left({\left(x - 2\right)}^{\frac{3}{4}}\right)}^{\frac{4}{3}} = {8}^{\frac{4}{3}}$

${\left(x - 2\right)}^{1} = {\left({2}^{3}\right)}^{\frac{4}{3}}$

$x - 2 = {2}^{4} = 16$

$x = 18$,

But we can get other solutions:

${\left(x - 2\right)}^{\frac{3}{4}} = 8$

${\left({\left(x - 2\right)}^{\frac{3}{4}}\right)}^{4} = {8}^{4}$ This step is dangerous since we can be adding false solutions.

${\left(x - 2\right)}^{3} = {8}^{4} = 4096$

${x}^{3} - 6 {x}^{2} + 12 x - 8 = 4096$

${x}^{3} - 6 {x}^{2} + 12 x - 4104 = 0$

We already know that 18 is one of the zeros:

So we divide the third degree polynom by (x-18):

$\frac{{x}^{3} - 6 {x}^{2} + 12 x - 4104}{x - 18} = {x}^{2} + 12 x + 228$

${x}^{2} + 12 x + 228$ can be solved by the quadratic formula:

$x = \frac{- 12 \pm \sqrt{{12}^{2} - 4 \cdot 228}}{2}$

$x = \frac{- 12 \pm \sqrt{- 768}}{2}$

$x = \frac{- 12 \pm 16 \sqrt{- 3}}{2}$

$x = - 6 \pm 8 i \sqrt{3}$

These are false solutions, since ${\left(x - 2\right)}^{\frac{3}{4}} = \pm 8 i$. These false solutions were added in the step where it is written :"This step is dangerous since we can be adding false solutions."

So the only solution is x=18.

Mar 25, 2016

$x = 18$

Assuming no complex solution is required

#### Explanation:

Given:$\text{ } {\left(x - 2\right)}^{\textcolor{m a \ge n t a}{\frac{3}{4}}} = 8$

Remember that ${\left(x - 2\right)}^{\frac{3}{4}} = \sqrt[3]{{\left(x - 2\right)}^{4}}$

Take the cube root of both sides giving

$\text{ } {\left(x - 2\right)}^{\textcolor{m a \ge n t a}{\frac{1}{4}}} = 2$

Raise both sides to the power of 4 giving

$x - 2 = {2}^{4} = 16$

$x = 18$