(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.
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
1/3 x 3/4 = 1/4
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.
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
(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
-1
-21
(parentheses), {braces}, and [brackets] EXAMPLE: for the nubmers 1 x 3 + 2 x 4 (1 x 3) + (2 x 4) = 3 + 8 = 11 1 x (3 + 2) x 4 = 1 x 5 x 4 = 20