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x^(-5/3) = 4 1/x^(5/3) = 4 x^(5/3) = 1/4 [x^(5/3)]^(3/5) = (1/4)^(3/5) x = 1/4^(3/5) or, x = 4^(-3/5) Check: x^(-5/3) = 4 [4^(-3/5)]^(-5/3) = 4 ? 4 ^(15/15) = 4 ? 4^1 = 4 ? 4 = 4
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(x^2+x-1/2)= x(x+1)-1/2 [x + (1 - square root of 3)/2][x + (1 + square root of 3)/2] = 0 Check it: x^2 + x/2 + (square root of 3)x)/2 + x/2 + 1/4 + (square root of 3)/4 - (square root of 3)x/2 - (square root of 3)/4 - 3/4 = 0 x^2 + x/2 + x/2 + [(square root of 3)x]/2 - [(square root of 3)x]/2 + (square root of 3)/4 - (square root of 3)/4 + 1/4 - 3/4 = 0 x^2 + x - 2/4 = 0 x^2 + x - 1/2 = 0 How to find this roots: Using the completing the square method: x^2 + x - 1/2 = 0 x^2 + x = 1/2 x^2 + x + 1/4 = 1/2 + 1/4 (x + 1/2)^2 = 3/4 x + 1/2 = (plus & minus)(square root of 3/4) x = -1/2 + (square root of 3)/2 x = - 1/2 - (square root of 3)/2
So, we have 3x = 4 for some number x. We will solve for x. 3x = 4 (1/3)*(3x) = (1/3)*(4) [multiplied by (1/3) on both sides] x = 4/3
x + 1 = y y + 3 = z z = y + 3 = (x + 1) + 3 = x + 4 Or: x = y - 1 = (z - 3) - 1 = z - 4 Which results in the same: x exceeds z by 4.