11x(2x - 1) = 0 so either 11x = 0 or x = ½
4x2 - x - 5 = 0 4x2 + 4x - 5x - 5 = 0 4x(x + 1) - 5(x + 1) = 0 (4x - 5)(x + 1) = 0 4x - 5 = 0 or x + 1 = 0 4x = 5 or x = -1 x = 5/4 o x = -1
8 (eight)
'x' must be zero, so the solution set is [ 0 ].
There are four solutions to 9x4 plus 27x3 - 4x2 - 12x equals 0. One of the solutions is x = -3.
7 and (-3/2)
y + 5x = 6 if x = {-1, 0, 1}
By using the quadratic equation formula: (4x+1)(x-3) = 0
x2 + 11x + 18 (x + 9)(x + 2) CHECK: x2 + 9x + 2x + 18 x2 + 11x + 18 SET EACH EQUAL TO ZERO: x + 9 = 0 x = -9 x + 2 = 0 x = -2 NOW YOU ARE DONE: Solution set: {-9, -2}
17x2-11x+2 = 2x2 Subtract 2x2 from both sides: 15x2-11x+2 = 0 Factorise: (3x-1)(5x-2) = 0 Solution: x =1/3 or x = 2/5
4x2-4x-3 = 0 (2x-3)(2x+1) = 0 x = 3/2 or x = -1/2
11x(2x - 1) = 0 so either 11x = 0 or x = ½
4x2 - 2x = 0
x2 = -11x - 10 x2 + 11x + 10 = 0 (x + 1)(x + 10) = 0
To solve for x: x2 = 11x - 10 x2 - 11x = -10 x2 - 11x + 10 =0 (x - 1)(x - 10) = 0 x = {1, 10)
4x2 - 4x + 1 = 0 => (2x - 1)(2x - 1) = 0 => (2x - 1)2 = 0 There is one solution: x = 1/2. It is a repeated root of the equation.
x2 = 11x - 10 ∴ x2 - 11x + 10 = 0 ∴ (x - 10)(x - 1) = 0 ∴ x ∈ {1, 10}