Using this substitution the equation becomes,

\[{u^2} - 6u + 7 = 0\]

Now, this equation does not factor. That happens on occasion but luckily enough we know how to solve it anyway. All we need to do is use the quadratic formula to find the solutions to this equation.

\[u = \frac{{ - \left( { - 6} \right) \pm \sqrt {{{\left( { - 6} \right)}^2} - 4\left( 1 \right)\left( 7 \right)} }}{{2\left( 1 \right)}} = \frac{{6 \pm \sqrt 8 }}{2} = \frac{{6 \pm 2\sqrt 2 }}{2} = 3 \pm \sqrt 2 \]

Do not get excited about the “messy” numbers here! These kinds of solutions happen on occasion and we just need to be able to deal with them. Just keep in mind that they are just numbers even if they are not the integers we are used to seeing!

Show Step 3

Now all we need to do is use our substitution from the first step to determine the solution to the original equation.

\[u = 3 + \sqrt 2 :\hspace{0.25in}{x^4} = 3 + \sqrt 2 \hspace{0.25in} \Rightarrow \hspace{0.25in}x = {\left( {3 + \sqrt 2 } \right)^{\frac{1}{4}}} = {\left( {4.4142} \right)^{\frac{1}{4}}} = 1.4495\] \[u = 3 - \sqrt 2 :\hspace{0.25in}{x^4} = 3 - \sqrt 2 \hspace{0.25in} \Rightarrow \hspace{0.25in}x = {\left( {3 - \sqrt 2 } \right)^{\frac{1}{4}}} = {\left( {1.5858} \right)^{\frac{1}{4}}} = 1.1222\]

Therefore the two solutions to the original equation are : \[\require{bbox} \bbox[2pt,border:1px solid black]{{x = 1.1222\,\,{\mbox{and }}x = 1.4495}}\] .