Solve this equation. (2x^3-2x^2-24x/2x^2-8x) = -1
Okay so, we take 2x common out from both numerator and denominator, so: 2x(x^2-x-12)/2x(x-4) = -1 Cancel 2x from both numerator and denominator so we get x^2-x-12/x-4 = -1 Take x-4 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.
Many thanks! Now I know how to solve these type of problems.
No problem bro! :D
8D
Im kind of struggling with this one, think you could help? ^^ (5x^3+20x^2-105x/5x^2-15x) = 0
Let's see.. Take 5x out common from numerator and denominator and cancel it. We now have: x^2 +4x -21/x-3 =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)(x-3) = 0 x = -7 or x= 3
Brilliant! kk so I was on the right track xD Thanks again dude
Always a pleasure.
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