# How do you solve root4(x^4+1)=3x?

Jul 31, 2016

$x = \frac{1}{2} \sqrt[4]{\frac{1}{5}}$

#### Explanation:

$\sqrt[4]{{x}^{4} + 1} = 3 x$
or
${x}^{4} + 1 = {\left(3 x\right)}^{4}$
or
${x}^{4} + 1 = 81 {x}^{4}$
or
$81 {x}^{4} - {x}^{4} = 1$
or
$80 {x}^{4} = 1$
or
${x}^{4} = \frac{1}{80}$
or
$x = \sqrt[4]{\frac{1}{80}}$
or
x=root4(1/((16)(5))
or
$x = \frac{1}{2} \sqrt[4]{\frac{1}{5}}$

Jul 31, 2016

$x = \frac{1}{2 \sqrt[4]{5}}$

#### Explanation:

root(4)(x^4+1)=3x ?

${\left(\sqrt[4]{{x}^{4} + 1}\right)}^{4} = {\left(3 x\right)}^{4}$

${x}^{4} + 1 = 81 {x}^{4}$

$1 = 81 {x}^{4} - {x}^{4}$

$1 = 80 {x}^{4}$

$\sqrt[4]{1} = \sqrt[4]{80 {x}^{4}}$

$1 = x \sqrt[4]{{2}^{4} \cdot 5}$

$1 = 2 x \sqrt[4]{5}$

$x = \frac{1}{2 \sqrt[4]{5}}$

Jul 31, 2016

A trick to help solve roots

#### Explanation:

If you are not sure how to deal with big number roots build a prime factor tree. In the question you are looking for something you can obtain a whole number 4th root from. You mat only be able to do it for part of the number leaving something behind in the root. As in this case.

$\textcolor{b l u e}{\text{Prime factor tree of 80 - looking for 4th root}}$

So $\sqrt{80}$ is correct and so is $\sqrt{{2}^{2} \times 5} = 2 \sqrt{5}$