## Simplifying (I'll stop, I promise)

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
EdSewersPark
Posts: 22
Joined: Mon May 23, 2011 8:30 pm
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### Simplifying (I'll stop, I promise)

OK, so where am I going wrong here?

Simplify: $(\sqrt[4]{2}+\sqrt[4]{8})^2$

So I converted $\sqrt[4]{x}$ into $x^{\fra{1}{4}}$

$(2^{\frac{1}{4}}+8^{\frac{1}{4}})$$(2^{\frac{1}{4}}+8^{\frac{1}{4}})$ =

$(4^{\frac{2}{4}}+16^{\frac{2}{4}}+16^{\frac{2}{4}}+64^{\frac{2}{4}})$ =

$(\sqrt[4]{16}+\sqrt[4]{256}+\sqrt[4]{256}+\sqrt[4]{4096})$ =

$(2+4+4+8)=18$

But I'm told the answer is: $4+3\sqrt{2}$ or $4+3\sqrt[4]{4}$ so, obviously I cannot do the conversion (why?)

So I did it without the conversion:

Simplify: $(\sqrt[4]{2}+\sqrt[4]{8})^2$

$(\sqrt[4]{2}+\sqrt[4]{8})(\sqrt[4]{2}+\sqrt[4]{8})$ =

$(\sqrt[4]{4}+\sqrt[4]{16}+\sqrt[4]{16}+\sqrt[4]{64})$ =

and here's where I'm lost: $(\sqrt[4]{4}+2+2+\sqrt[4]{64})$ = $(4+\sqrt[4]{4}+\sqrt[4]{64})$

Luke53
Posts: 54
Joined: Sun Mar 13, 2011 9:46 am
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### Re: Simplifying (I'll stop, I promise)

Simplify: ( 2^(1/4) + 8^!1/4))^2
(2^(1/4))^2 = 4^(1/4) = sqrt (2)
16^(1/4) = 2 (twice)
64^(1/4) =sqrt 8 = 2*sqrt (2)
the sum is: sqrt (2) + 2 + 2 + 2 * sqrt(2) = 4 + 3 * sqrt (2).

EdSewersPark
Posts: 22
Joined: Mon May 23, 2011 8:30 pm
Contact:

### Re: Simplifying (I'll stop, I promise)

Aha, these are the mistakes I make every time.