If you mean: x2+5x+6 = 0 then the solutions are x = -3 and x = -2
(x-2)(x-4) = 0 x = 2 or x = 4
x2 - 6x + 8 = 0 x2 - 4x - 2x + 8 = 0 x*(x - 4) - 2*(x - 4) = 0 (x - 2)*(x - 4) = 0 so (x - 2) = 0 or (x - 4) = 0 ie x = 2 or x = 4
y = 0x = y + 2plug 0 in for yx = 0 + 2x = 2
2x2+5x-3 = 0 (2x-1)(x+3) = 0 x = 1/2 or x -3
If you mean: x2+5x+6 = 0 then the solutions are x = -3 and x = -2
x2+x-6=0 * * * * * Perhaps the solution you are looking for is: x2 + x - 6 = (x - 2)(x + 3) = 0; whence, x = 2 or -3.
first we assume this to be true. x+1=x+2 x=x+1 x-x=1 0=1 this is not true, so the initial premise that this problem has a solution must be false. There is no solution.
(x-2)(x-4) = 0 x = 2 or x = 4
x2 - 5x + 6 = 0(x - 2) (x - 3) = 0x - 2 = 0 5 × 69 or x - 3 = 0x = 2 or x = 3
If x - 4y = 2 and x + 4y = 2 then the only solution is when y = 0 and thus x = 2.
x2 - 6x + 8 = 0 x2 - 4x - 2x + 8 = 0 x*(x - 4) - 2*(x - 4) = 0 (x - 2)*(x - 4) = 0 so (x - 2) = 0 or (x - 4) = 0 ie x = 2 or x = 4
2x2-7x+3 = 0 (2x-1)(x-3) = 0 x = 1/2 or x = 3
y = 0x = y + 2plug 0 in for yx = 0 + 2x = 2
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
x2 + 6x = 16x2 + 6x - 16 = 0(x + 8)(x - 2) = 0x + 8 = 0 or x - 2 = 0x = -8 or x = 2Since it is given that x > 0, then x = -8 is not a solution of the equation. Thus, the only solution is x = 2.
2x2+5x-3 = 0 (2x-1)(x+3) = 0 x = 1/2 or x -3