This is a quadratic equation question in finding the possible values of x x2 - 6x = - 8 x2 - 6x + 8 = 0 Factorise the expression in the equation: (x-2)(x-4) = 0 Therefore: x = 2 or x = 4
If you mean: x2+6x+8 then it is (x+4)(x+2) when factored
In the equation x2 = 6x - 9, all terms must be moved to one side of the equals sign, giving x2 - 6x + 9 = 0. This becomes factorable to (x -3)(x-3).
If -6x = 8, then x = -4/3.
x2 + 6x + 8 =(x + 2)(x +4)
2x2 + 6x - 8 = 72 ∴ 2x2 + 6x + 64 =0 ∴ x2 + 3x + 32 = 0 This can not be factored, as x is not equal to any integer. Using the quadratic equation, we find that: x = -3/2 ± √119 / 2i
The quadratic expression x2+6x+8 when factorised equals (x+2)(x+4)
x2 + 6x = 16=> x2 + 6x - 16 = 0=> x2 + 8x -2x - 16 = 0=> (x+8)(x-2) = 0=> x = -8 or x = 2So, the solutions of the quadratic equation x2 + 6x = 16 are -8 and 2.
8
2x+6x=-9 => 8x=-9=> x=-8/9
x2 + 6x + 8 = 0 Solve for x.X = -2 or X = -4
-x2 + 6x + 16 = -(x2 - 6x - 16) = -(x - 8)(x + 2) = -(8 - x)(x + 2)
x2-6x+8 = (x-2)(x-4) when factored
x2 - 6x - 16 = (x - 8)(x + 2)
x2 + 6x + 8 = (x + 4)(x + 2)
x2 + 6x + 1 = 0 x2 + 6x + 9 = 8 (x + 3)2 = 8 x + 3 = ± 2√2 x + 3 = -3 ± 2√2 x ∈ {-3 - 2√2, -3 + 2√2}
If you mean: x2+6x+8 then it is (x+4)(x+2) when factored
x2 + 6x + 9 = 81 x2 + 6x = 72 x2 + 6x - 72 = 0 (x+12)(x-6) = 0 x= -12, 6 (two solutions)