$x^{\frac{1}{2}}-3x^{\frac{1}{4}}-10=0$

Question

$$x^{\frac{1}{2}}-3x^{\frac{1}{4}}-10=0$$ $$t=x^{\frac{1}{4}}$$ $$t^2-3t-10=0$$ $$\Delta=49$$ $$t_1=5$$ $$t_2=-2$$ And now I'm not sure what should I do... $$x_1^{\frac{1}{4}}=5$$ $$x_1=5^4=625$$ I'm not sure why did it work, but that's correct the answer. But what with $x_2$? $$x_2^{\frac{1}{4}}=-2 $$ $$x_2^{\frac{4}{4}}=(-2)^{4}$$ $$x_2=16$$ My book clearly says that there's only one answer : $625$. So why is the $x_2=16$ wrong? :(

Answer 1

This is not a full answer to the question but it is too long for a comment.

Consider function and derivatives $$f(x)=\sqrt{x}-3 \sqrt[4]{x}-10$$ $$f'(x)=\frac{1}{2 \sqrt{x}}-\frac{3}{4 x^{3/4}}$$ $$f''(x)=\frac{9}{16 x^{7/4}}-\frac{1}{4 x^{3/2}}$$ The first derivative cancels only at $x=\frac{81}{16}$ (it is negative before, positive after). For this value $$f(\frac{81}{16})=-\frac{49}{4}$$ $$f''(\frac{81}{16})=\frac{8}{729}$$ so the point corresponds to a minimum. Since, moreover, $f(0)=-10$, there is only one root to the equation (which you found).