Solve the equation $x^2-9\sqrt{x}+14=0$

Question

Solve the equation $$x^2-9\sqrt{x}+14=0$$

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I think there are no real solutions. How to prove it?

Answer 1

Let $\sqrt{x}=u$ then let $$u^4-9u+14=f(u)$$

From which one must show that $f(u)$ is positive for $u \geq 0 $

Answer 2

Let's solve the equation $$t^4-9t+14=0$$ The minimum value for the polynomial occurs when $t=\sqrt[3]{9/4}$ and it's $$\sqrt[3]{2.25^4}-9\sqrt[3]{2.25}+14>\sqrt[3]{16}-9\sqrt[3]3+14>2.5-13+14>0$$

So it has no solution.

Remarks:

$$2.5^3=125/8<128/8=16$$ $$13^3=2197>2187=3^7=9^3\cdot 3$$