Prove this

Question

Prove this

sasogeek · Answer

attachment

sasogeek · Answer

when i try to simplify the right hand side, i get log2$\sqrt{ab}$ ... i'm not sure how to prove the truth of the equation but it looks like it requires a lot of thinking which i'm not doing... it's most likely something simple too... i need help here

zarkon · Answer

$$0=\frac{1}{3}(a^2-(a^2+b^2)+b^2)$$
$$=\frac{1}{3}(a^2-14ab+b^2)$$

$$=\frac{1}{3}(a^2-2ab+b^2)-\frac{1}{3}12ab$$

$$=\frac{1}{3}(a^2-2ab+b^2)-4ab$$

$$=\frac{1}{3}(a-b)^2-4ab$$

thus 

$$\frac{1}{3}(a-b)^2=4ab$$


$$\frac{1}{\sqrt{3}}(a-b)=2\sqrt{ab}$$

sasogeek · Answer

ummm thanks but could u please explain ur first statement? i think i can follow from there... where did the 0 and... actually how did that equation come about?

zarkon · Answer

the first equation is just a fact

$$\frac{1}{3}(a^2-(a^2+b^2)+b^2)=\frac{1}{3}(a^2-a^2-b^2+b^2)$$

$$=\frac{1}{3}((a^2-a^2)+(-b^2+b^2))=\frac{1}{3}(0+0)=0$$