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anonymous
 4 years ago
Solve this equation.
(2x^32x^224x/2x^28x) = 1
anonymous
 4 years ago
Solve this equation. (2x^32x^224x/2x^28x) = 1

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay so, we take 2x common out from both numerator and denominator, so: 2x(x^2x12)/2x(x4) = 1 Cancel 2x from both numerator and denominator so we get x^2x12/x4 = 1 Take x4 to the other side and multiply by 1. x^2 x 12 =x +4 Cancel x from both sides since it is common. x^2 12 = 4 Take 12 to other side, x^2 = 4+12 x^2 = 16 Or by taking square root: x= 4 is the answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Many thanks! Now I know how to solve these type of problems.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Im kind of struggling with this one, think you could help? ^^ (5x^3+20x^2105x/5x^215x) = 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let's see.. Take 5x out common from numerator and denominator and cancel it. We now have: x^2 +4x 21/x3 =0 Remove the denominator because anything multiplied by 0 is 0. So, x^2 +4x 21 = 0 x^2 +7x 3x 21= 0 (x+7)(x3) = 0 x = 7 or x= 3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Brilliant! kk so I was on the right track xD Thanks again dude
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