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

Question

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 05:46 — Inceptio, Mar 28 '13 at 05:46
Hint: Denote $u=\sqrt[4]{x}$, you will get $\sqrt{x}=u^2$. – gev Mar 28 '13 at 05:47 — gev, Mar 28 '13 at 05:47

lsp · Accepted Answer · 2014-01-09T06:39:49.970

6

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 $x = 4^4 = 256$

edited Jan 09 '14 at 06:39

answered Mar 28 '13 at 05:54

lsp

4,745

Stahl · Answer 2 · 2013-03-28T05:57:01.503

2

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).

edited Mar 28 '13 at 05:57

answered Mar 28 '13 at 05:48

Stahl

23,212

score 1 · Answer 3 · edited Mar 28 '13 at 18:38

1

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.

edited Mar 28 '13 at 18:38

MJD

65,394
39
298
580

answered Mar 28 '13 at 05:48

Inceptio

7,881

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

3 Answers3