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
3
(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.
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
(x - 3)/(12 - 4x) = (x - 3)/[4(3 - x)] = (x - 3)/[-4(x - 3)] = 1/-4 = -(1/4)
3/4 x 4/5 = 12/20 = 6/10 = 3/5 or three fifthsNotice that the fours cancel out. Think about writing it as:3 x (1/4) x 4 x (1/5) = 3 x [(1/4) x 4] x (1/5) = 3 x [1] x (1/5) = 3 x (1/5) = 3/5.
1 = (1 + 4) ÷ (2 + 3) 2 = 4 - 3 + 2 - 1 3 = (4 + 3 - 1) ÷ 2 4 = (4 + 3 + 1) ÷ 2 5 = (4 x 2 - 3) x 1 6 = 4 + 3 - 2 + 1 7 = (4 + 3) x (2 - 1) 8 = 4 + 3 + 2 - 1 9 = 4 + 3 + 2 x 1 10 = 4 + 3 + 2 + 1 11 = 4 x 2 + 3 x 1 12 = 4 x 3 x (2 - 1) 13 = 3 x 4 + 2 - 1 14 = 3 x 4 + 2 x 1 15 = 3 x 4 + 2 + 1
3
x2 + x + 1 = 0 ∴ x2 + x + 1/4 = -3/4 ∴ (x + 1/2)2 = -3/4 ∴ x + 1/2 = ± √(-3/4) ∴ x = - 1/2 ± (i√3) / 2 ∴ x = (-1 ± i√3) / 2
1
cube numbers next is 64 1 = 1 x 1 x 1 = 1³ 8 = 2 x 2 x 2 = 2³ 27 = 3 x 3 x 3 = 3³ then 64 = 4 x 4 x 4 = 4³
(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
1/3 x 3/4 = 1/4
1
Y = X+3, therefore: 3 1/2 4 1/2 5 1/2 7 1/4 7 3/4