I need help with verifying an algebraic equation.

Hedgehog

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Dec 22, 2010
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I have an extra credit assignment that I need some help with. I am currently in Trig but the problem is verifying that two algebraic equations are equal. I am rusty on my algebra -- one of the equations has a radical square root -- and I am having greAT DIFFICULTY. I have to verify that (square root of 2 + square root of 6) / 4 is equal to[the square root of (2 + the square root of 3)] /2. I can simplify the first part as the (square of 2)/4 plus the (square of 6)/4 but can not remember how the properties work to take it any further.
 
Hello, Hedgehog!

\(\displaystyle \text{Show that: }\:\frac{\sqrt{2} + \sqrt{6}}{4} \;=\;\frac {\sqrt{2 + \sqrt{3}}} {2}\)

\(\displaystyle \text{Let: }\,X \;=\;\frac{\sqrt{2} + \sqrt{6}}{4}\)

\(\displaystyle \text{Square: }\:X^2 \;=\;\left(\frac{\sqrt{2} + \sqrt{6}}{4}\right)^2 \;=\;\frac{2 + 2\sqrt{12} + 6}{16} \;=\; \frac{8 + 4\sqrt{3}}{16}\)

\(\displaystyle \text{Hence: }\:X^2 \;=\; \frac{2 + \sqrt{3}}{4} \quad\Rightarrow\quad X \;=\;\sqrt{\frac{2 + \sqrt{3}}{4}} \;=\;\frac{\sqrt{2+\sqrt{3}}}{2}\)


\(\displaystyle \text{Therefore: }\:\frac{\sqrt{2}+\sqrt{6}}{4} \;=\;\frac{\sqrt{2+\sqrt{3}}}{2}\)

 
Hedgehog & lookagain edit said:
...the problem is verifying that two algebraic \(\displaystyle expressions\) are equal. //... -- one of the \(\displaystyle expressions\)
has a radical square root -- .... I have to verify that (square root of 2 + square root of 6) / 4 is equal to
the square root of (2 + the square root of 3)] /2.

Hedgehog,

you are talking about algebraic expressions, not algebraic equations, as I amended part of
your post in the above quote box to reflect that.

You are to be verifying that two algebraic expressions are equal to each other.
 
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