Algebraic expression #6, #7

[mathwannabe]

Hello everyone

Couldn't sleep so I figured it would be useful to do some mathematics instead of watching a film or looking at the ceiling.

I tried to simplify some expressions that involve radicals. I am just unsure about the answer I got with one of them:

1) \(\displaystyle (1+[\frac{1}{2}(\frac{b}{a})^{-\frac{1}{2}}-\frac{1}{2}(\frac{a}{b})^{-\frac{1}{2}}]^{-2})^{\frac{1}{2}}=\)

\(\displaystyle =(1+[\frac{1}{2}\sqrt{\frac{a}{b}}-\frac{1}{2}\sqrt{\frac{b}{a}}]^{-2})^{\frac{1}{2}}=\)

\(\displaystyle =(1+[\frac{\sqrt{a}}{2\sqrt{b}}-\frac{\sqrt{b}}{2\sqrt{a}}]^{-2})^{\frac{1}{2}=\)

\(\displaystyle =(1+[\frac{a-b}{2\sqrt{ab}}]^{-2})^{\frac{1}{2}})=\)

\(\displaystyle =(1+ \frac{4ab}{(a-b)^2})^{\frac{1}{2}=\)

\(\displaystyle =(\frac{(a-b)^2}{(a-b)^2}+\frac{4ab}{(a-b)^2)})^{\frac{1}{2}}=\)

\(\displaystyle =(\frac{(a-b)^2+4ab}{(a-b)^2})^{\frac{1}{2}=\)

\(\displaystyle =(\frac{(a+b)^2}{(a-b)^2})^{\frac{1}{2}=\)

\(\displaystyle =\sqrt{\frac{(a+b)^2}{(a-b)^2}}=\frac{a+b}{a-b}\) >> But, I am unsure if the answer should actually be >> \(\displaystyle |\frac{a+b}{a-b}|\) ?

 

[mathwannabe]

\(\displaystyle For\ any\ real\ number\ x, \sqrt{x^2} = |x|.\)

That is the radical implies non-negativity. So \(\displaystyle \sqrt{(-3)^2} = |-3| = 3.\)

Hello Jeff, thank you for the reply.