So I saw this math problem and I don't really know how to solve it... I'm learning quadratic functions but this one is not in the form ax^2+bx+c so not sure how to solve. Anyways... (Don't remember exact equation so will not be exact) 2x / x + 5 = 3 / x^2-8 There might be no possible solutions since i just made it up. I would just like to know how to solve this type of equation, if this example doesn't work, it would be nice if u guys solved it step by step with something that works? Thanks!
2x / x + 5 = 3 / x^2-8 Multiply all terms by (x + 5)(x^2 - 8): (2x)(x + 5)(x^2 - 8)/(x + 5) = (3) (x + 5)(x^2 - 8)/(x^2 - 8) Cancel where possible: 2x(x^2 - 8) = 3(x + 5) 2x^3 - 16x = 3x + 15 2x^3 - 19x - 15 = 0 x(2x^2 - 19) - 15 = 0 1. x - 15 = 0 x = 15 2. 2x^2 - 19 = 0 2x^2 = 19 x^2 = 19/2 x = the square root of 19/2 There may be a way of solving problem further, but I'm not familiar with it.
This problem would work for the quadratic formula if the x^2 was changed to x, so I will show you what it would look like and how to solve it. (2x)/(x+5)=3/(x-8) To solve this kind of problem, first you cross multiply, which means that you would multiply the denominators to the opposite side, so it would look like: (2x)(x-8)=3(x+5) Then you distribute (multiply it out) to get: 2x^2-16x=3x+15 Move everything to one side to make it equal to zero: 2x^2-16x-3x-15=0 Add the two x's together: 2x^2-19x-15=0 Now you can use the quadratic formula!!!! (: