Answer
$m=81$ or $m=16$
Work Step by Step
$ m^{\frac{1}{2}}+6=5m^{\frac{1}{4}}\qquad$... add $-5m^{\frac{1}{4}}$ to both sides.
$ m^{\frac{1}{2}}-5m^{\frac{1}{4}}+6=0\qquad$...substitute $m^{\frac{1}{4}}$ for $u$ so that $u^{2}=m^{\frac{1}{2}}$
Note that $m^{\frac{1}{4}}$ is the fourth root of $t$, a positive number.
$ u^{2}-5u+6=0\qquad$... solve with the Quadratic formula.
$u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$u=\displaystyle \frac{5\pm\sqrt{25-24}}{2}$
$u=\displaystyle \frac{5\pm\sqrt{1}}{2}$
$u=\displaystyle \frac{5\pm 1}{2}$
$u=\displaystyle \frac{5+1}{2}=\frac{6}{2}=3$ or $u=\displaystyle \frac{5-1}{2}=\frac{4}{2}=2$
Bring back $m^{\frac{1}{4}}=u$.
$m^{\frac{1}{4}}=3$ or $ m^{\frac{1}{4}}=2\qquad$...raise both sides of both expressions to the fourth power.
$(m^{\frac{1}{4}})^{4}=3^{4}$ or $(m^{\frac{1}{4}})^{4}=2^{4} $
$m=81$ or $m=16$
