2x-3=4x+9 -2x-12=0 2x=-12 x=-6
4x+2y-8=0 2y=8-4x y=4-2x slope= -2
Given, 4x2 - 4x + 3 = 0; then, 4x2 - 4x + 1 = (2x - 1)2 = -2, 2x - 1 = √-2 = ±i√2, and 2x = 1 ± i√2; therefore, x = ½(1 ± i√2).
If: y = x^2 -2x +4 and y = 2x^2 -4x +4 Then: 2x^2 -4x +4 = x^2 -2x +4 Transposing terms: x^2 -2x = 0 Factorizing: (x-2)(x+0) => x = 2 or x = 0 Therefore by substitution points of intersect are at: (2, 4) and (0, 4)
If -2x +14 = 0, -2x = -14, or x = 7.
2x + 2x + 146 + 26 = 172 4x + 172 = 172 4x = 0 x = 0
many solutions
2x-3=4x+9 -2x-12=0 2x=-12 x=-6
(2x + 1)(2x + 1) or (2x + 1)2 so x = -0.5
2x = x2 + 4x - 3 x2 + 2x - 3 = 0 (x - 1)(x - 2) = 0
(4x + 3)(2x - 1) so x = -0.75 or 0.5
4x+2y-8=0 2y=8-4x y=4-2x slope= -2
Given, 4x2 - 4x + 3 = 0; then, 4x2 - 4x + 1 = (2x - 1)2 = -2, 2x - 1 = √-2 = ±i√2, and 2x = 1 ± i√2; therefore, x = ½(1 ± i√2).
If: y = x^2 -2x +4 and y = 2x^2 -4x +4 Then: 2x^2 -4x +4 = x^2 -2x +4 Transposing terms: x^2 -2x = 0 Factorizing: (x-2)(x+0) => x = 2 or x = 0 Therefore by substitution points of intersect are at: (2, 4) and (0, 4)
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.
8x2-2x-3=0 (2x+1)(4x-3)=0 2x+1=0 or 4x-3=0 x=-1/2 or x=3/4
It is a contradiction! 6x - 4x - 6 = 2x 2x - 6 = 2x -6 = 0 false