Use a calculator to find the value of the expression. Then show that the calculated value is correct. frac{sqrt{2}+sqrt{6}}{sqrt{2+sqrt{3}}} =2 ~~I have tried subbing in a common variable to prove this, such as m= sqrt{2} and using that, put it into the equation so that the equation now reads: frac{m+(m^2+4)^{1/2}}{sqrt{2+(m^2+1)^{1/2}}} I have not gotten very far with that though, I may be doing it completely wrong.So any insight/help is appreciated! Thanks!
\[\frac{\sqrt{2}+\sqrt{6}}{\sqrt{2+\sqrt{3}}} =2\] is the original equation.
\[\frac{m+(m^2+4)^{1/2}}{\sqrt{2+(m^2+1)^{1/2}}}\] This is the equation I've gotten when I sub in m=\[\sqrt{2}\]
First, square original equation and show me what u get...
Okay... \[\frac{2+2(2^{1/2})(6^{1/2})+6}{2+\sqrt{3}} =4\] Thats what I get when I square both sides of the original equation.
That's right, now can u break down sqrt 6 ?
Do you mean combine 2^1/2 and 6^1/2 which is 12^1/2?
That's true but I meant just factor sqrt 6 into 2 square roots.
Okay so sqrt6= sqrt2*sqrt3 and then when you put that in the original equation that's all times the sqrt2 so it the numerator becomes 8+4*sqrt3..?
When you squared, you got: \[\frac{2+2(2^{1/2})(6^{1/2})+6}{2+\sqrt{3}} =4\] Now substitute sqrt6= sqrt2*sqrt3 into that and simplify.
Ohhh!!!! Okay! I got it!! Thank you so much for always being so helpful! :)
Well done. No need for algebra to solve radicals (usually), squaring once (or maybe twice) usually works.
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