SOLUTION: 2 SQRT over6^2 + 6 SQRT over 2^2 = c^2 My friend said that this was the answer.. 12 + 12 = c^2 c=SQRT over 24 But I don't know where she got it or if its right. I'm really

Algebra ->  Radicals -> SOLUTION: 2 SQRT over6^2 + 6 SQRT over 2^2 = c^2 My friend said that this was the answer.. 12 + 12 = c^2 c=SQRT over 24 But I don't know where she got it or if its right. I'm really       Log On


   

Question 212275: 2 SQRT over6^2 + 6 SQRT over 2^2 = c^2
My friend said that this was the answer..
12 + 12 = c^2
c=SQRT over 24
But I don't know where she got it or if its right.
I'm really not good with Square roots.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Is the problem 2%2Asqrt%286%5E2%29%2B6%2Asqrt%282%5E2%29=c%5E2 ???


2%2Asqrt%286%5E2%29%2B6%2Asqrt%282%5E2%29=c%5E2 Start with the given equation.


2%2A6%2B6%2Asqrt%282%5E2%29=c%5E2 Evaluate the square root of 6%5E2 to get 6.


2%2A6%2B6%2A2=c%5E2 Evaluate the square root of 2%5E2 to get 2.


12%2B12=c%5E2 Multiply


24=c%5E2 Combine like terms.


c%5E2=24 Rearrange the equation.


c=%22%22%2B-sqrt%2824%29 Take the square root of both sides.


c=sqrt%2824%29 or c=-sqrt%2824%29 Break up the 'plus/minus'


c=sqrt%284%2A6%29 or c=-sqrt%284%2A6%29 Factor 24 to get 4*6


c=sqrt%284%29%2Asqrt%286%29 or c=-sqrt%284%29%2Asqrt%286%29 Break up the square root.


c=2%2Asqrt%286%29 or c=-2%2Asqrt%286%29 Take the square root of 4 to get 2.


So the solutions are c=2%2Asqrt%286%29 or c=-2%2Asqrt%286%29