x^3y = 2, y = 4
Substitute 4 for y:
(x^3)(4) = 2
x^3 = 8
x = 2
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x**2-2x-8 Multiply first 2 terms: 1*-8=-8 Find factors that add up to second term: -2, 4 --> -2+4 = 2 -4,-2 --> -4+2 = -2 Rewrite: x**2+2x-4x-8 Factor sets of terms x**2+2x=x(x+2) -4x-8=-4(x+2) x**2+2x-4x-8 = x(x+2)-4(x+2) Factor x(x+2)-4(x+2) = (x+2)(x-4)
x-2(x)+4/x^2 -4=x-2x+4/x^2 -4=-x-4+4/x^2
Don't know how you find it but I guess the answer is -2. -2 x -2 =4 4 X -2 = -8
(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
2^(6) = 64 2 x 2 x 2 x 2 x 2 x2 = 4 x 4 x 4 = 16 x 4 = 64