Answer
$x = 16$
Work Step by Step
Let $y = x^\frac{1}{4}$. We can re-write the original equation: $$x^\frac{1}{2} + 3x^\frac{1}{4} - 10 = 0$$ as so: $$(x^\frac{1}{4})^2 + 3(x^\frac{1}{4}) - 10 = 0$$ $$y^2 + 3y - 10 = 0$$ which is a quadratic equation. This can now be factorized by finding two factors of 10 that, when subtracted, result in $+3$: $$(y+5)(y-2)=0$$ where $$y + 5 = 0$$ $$y = -5$$ $$OR$$ $$y - 2=0$$ $$y=2$$ Since we'd already established taht $y = x^\frac{1}{4}$, we do the appropriate substitution in each case: $$y = -5$$ $$x^\frac{1}{4} = -5$$ $$(x^\frac{1}{4})^4 = (-5)^4$$ $$x = 625$$ $$OR$$ $$y=2$$ $$(x^\frac{1}{4})^4 = 2^4$$ $$x = 16$$ Substituting each value of $x$ into the original equation to verify each answer, we see the following: $$x^\frac{1}{2} + 3x^\frac{1}{4} - 10 = 0$$ $$(625)^\frac{1}{2} + 3(625)^\frac{1}{4} - 10 = 0$$ $$25 + 3(5) - 10 = 0$$ $$40 - 10 = 0$$ $$30\ne 0$$ $$OR$$ $$(16)^\frac{1}{2} + 3(16)^\frac{1}{4} - 10 = 0$$ $$4 + 3(2) - 10 = 0$$ $$4 + 6 - 10 = 0$$ $$0 = 0$$ We conclude that the only valid solution for the original equation is $x = 16$.
