Question

show that (2+root 2)^1/2 is rational

Answer 1

\(\sqrt{2+\sqrt{2}}\)? I wouldn't think so.

Answer 2

yes right x= thart

Answer 3

its correct what did you wriite

Answer 4

Hello

Answer 5

If it's rational, we should be able to some up with two integers that reproduce it.

\(\sqrt{2 + \sqrt{2}} = \dfrac{a}{b}\)

\(b^{2}\cdot (2 + \sqrt{2}) = a^{2}\)

\(b^{2}\sqrt{2} = a^{2} - 2b^{2}\)

\(b^{4}\cdot(2) = (a^{2} - 2b^{2})^{2} = a^{4} - 4a^{2}b^{2} + 4b^{4}\)

\(a^{4} - 4a^{2}b^{2} + 2b^{4} = 0\)

\(\dfrac{a^{4}}{b^{4}} - 4\dfrac{a^{2}}{b^{2}} + 2 = 0\)

\(\sqrt{(-4)^{2} - 4(1)(2)} = \sqrt{16 - 8} = \sqrt{8} = 2\sqrt{2}\)

Well, I think we just proved that a^2 / b^2 is not Rational (if we can assume sqrt(2) is not Rational.) That was a lot of work for an unconvincing argument.

Answer 6

I have other answer

Answer 7

do you want me to write it

Answer 8

its short

Answer 9

x^2-2=root 2
2=x^2-root2
x^4-4x^2+4=2
x^4-4x^2+2=0
x=+2,-2,+1,-1

Answer 10

my question is how I analysis equation from x^4

Answer 11

Substitute away.

\(2^{4} - 4(2)^{2} + 2 = 16 - 16 + 2 = 2 \ne 0\) - One Down.

Try the other three.

Answer 12

Very good, by the way. I didn't think of this simple application of the Rational Root Theorem.