# Solving $\sqrt{x}-2\sqrt[4]{x}-8 = 0$

How can I do this question? $$\sqrt{x}-2\sqrt[4]{x}-8 = 0$$

Can I solve this?

I tried to multiply everything by $x^4$, and got $$8x^4+x^3 -2x = 0$$

I don't know how to proceed from here.

• You don't get that if you multiply $x^{4}$ – Inceptio Mar 28 '13 at 5:46
• Hint: Denote $u=\sqrt[4]{x}$, you will get $\sqrt{x}=u^2$. – gev Mar 28 '13 at 5:47

Let $\sqrt[4]x=u$. This gives $\sqrt x = u^2$.

Solve the quadratic equation in $u$. You get the values for $u$ as $4$ or $-2$.

Since $\sqrt[4]x$ cannot be negative, it is equal to $4$.

Hence $x = 4^4 = 256$

When multiplying powers, we have to ADD exponents: $x^a x^b = x^{a + b}$. Instead, try letting $x = u^4$, so $$\sqrt{x} - 2\sqrt[4]{x} - 8 = 0$$ becomes $$u^2 - 2u - 8 = 0,$$ which you can solve using the quadratic formula (although you might introduce extraneous solutions).

Hint:

$x^{\frac{1}{4}}=k$

$\sqrt{(x)}=k^2$

Or you can just keep the equation as it is:

$\sqrt{x}-2\sqrt[4]{x}-8=0 \implies \sqrt{x}-4\sqrt[4]{x}+2\sqrt[4]{x}-8=0$. Factorize it.