# how do I solve this equation: 2000= (500 * (0.001/(3.14r^2)) + (0.0008/(3.14r^4), or even x^2+x^3=1100?

My big problem here is I don't kow how to equate powers. eg 2000= 500 ((x^2)^2) + x^4 how do I resolve for x. Or even x^2+x^3=1100. I know what x is as I made up the example but, if I hadn't made it up I suppose I would have to equate the x^2 and x^3 in some way and solve for that. Any help would be appreciated got an exam in 36 hours. thanks.

### 3 Answers

- 1 decade agoBest Answer
adding powers.. multiply the powers by each other.. e.g x^2+x^4=x^8 if this is then equalled to an amount, you find the root of that number to find x ..e.g x^2+x^4=1600 ... x^8=1600 ... bring the 8 over and find the root :-)

- gwennethLv 43 years ago
x2 – 4x – 5 = 0 (x - 5) (x + a million) = 0 So x = +5 or x = -a million For the quadratice equation: a = a million, b = -4, c = -5 Sqrt b^2 - 4ac = sqrt (sixteen - (4 * a million * -5)) = sqrt (sixteen - (-20)) = sqrt (36) = +6 or -6 The quadratic equation is then: [- (-4) +/- 6 ]/ 2 = 2 +/- 3 = +5 or -a million

- Anonymous1 decade ago
you should know that yourself