Help in solving this proof....Sqrt6+sqrt2/4=sqrt(inside is 2+sqrt3)/2?

Question

I tried squaring but it doesnt get me anywhere.

I need to prove that ( i will put -/ for square root)

-/6 + -/2 / 4 = -/(2+-/3) / 2


I hope that helps. remember that the 2+-/3 is inside of the square root. (It was hard to show without a square root key)

Answer 1

I'm going to assume that you mean the following question, as it is slightly ambiguous.

(√(6) + √(2)) / 4 = √[2 + √(3)] / 2

I've actually posted a similar question here a few days ago, if you're interested: http://answers.yahoo.com/question/index;_ylt=AnssBpjfs76T81sdgIe3vCLsy6IX?qid=20070331140906AAxDePv

Try to find a way to express 2 + √(3) as a perfect square.

2 + √(3) = (√a + √b)^2
2 + √(3) = a + 2√(ab) + b
2 + √(3) = (a + b) + 2√(ab)

Moving the 2 inside the √ , we get

2 + √(3) = (a + b) + √(4ab)

Now equate the rational parts and irrational parts.
a + b = 2
4ab = 3

Two equations, two unknowns. b = 2 - a, so
4a(2 - a) = 3
8a - 4a^2 = 3
0 = 4a^2 - 8a + 3
0 = 4a^2 - 6a - 2a + 3
0 = 2a(2a - 3) - (2a - 3)
0 = (2a - 1)(2a - 3)
Which means a = {1/2, 3/2}, b = {2 - (1/2), 2 - {3/2})

a = {1/2, 3/2} , b = {3/2, 1/2}

Therefore,
2 + √(3) = [√(a) + √(b)]^2
2 + √(3) = [√(1/2) + √(3/2)]^2

Next, plug this into the right hand side of the equation.

RHS = √(2 + √(3)) / 2
RHS = √[√(1/2) + √(3/2)]^2] / 2

But, the square root of a square is just itself, so the square root is eliminated.

RHS = [√(1/2) + √(3/2)]/2
RHS = [1/√(2) + √(3)/√(2)] / 2

Multiply top and bottom by √(2),

RHS = [1 + √(3)] / 2√(2)

Rationalize the denominator; multiply top and bottom by √(2),

RHS = [√2 + √2√3] / [2√2√2]

Simplify by merging square roots.

RHS = [√(2) + √(6)] / [2(2)]
RHS = [√(2) + √(6)] / 4 = LHS

Answer 2

Sounds like something that came out of a trig identity! Half-angle vs. angle sum?

(1) Multiply by 4 to get rid of the denominators.
-/6 + -/2 = 2*-/(2+-/3)

(2) Square both sides.
6 + 2 + 2sqrt(12) = 4*(2 + sqrt(3))

(3) Simplify...

Good luck!

P.S. You are missing a ( ) around sqrt(6) and sqrt(2) in the original statement.

Answer 3

You have a square root key. Apparently, you just don't know it. To produce the square root symbol, just hold down your "Alt" key while entering the number 251 on your ten-key keyboard. Here it is: √ . It is possible to produce all sorts of ASCII characters using your ten-key keyboard. Refer to the website below to find them:

http://www.asciitable.com/

You might want to put this in your favorites folder for easy access.