x2 + 5x + 3 = 9 x2 + 5x - 6 = 0 (x + 6)(x - 1) = 0 x = -6 or x = 1
We can't calculate what it equals until we know the value of ' x '.
x2 + 4x - 9 = 5x + 3 ∴ x2 - x - 12 = 0 ∴ (x + 3)(x - 4) = 0 ∴ x ∈ {-3, 4}
x2+4x-9 = 5x+3 x2+4x-5x-9-3 = 0 x2-x-12 = 0 (x+3)(x-4) = 0 x = -3 or x = 4
X2 - 5X = 24 X2 - 5X - 24 = 0 (X - 8)(X + 3) X = 8 ----------------and X = - 3 --------------
x2 + 5x + 3 = 9 x2 + 5x - 6 = 0 (x + 6)(x - 1) = 0 x = -6 or x = 1
x2-5x+4 = (x-1)(x-4) when factord
We can't calculate what it equals until we know the value of ' x '.
x2 + 4x - 9 = 5x + 3 ∴ x2 - x - 12 = 0 ∴ (x + 3)(x - 4) = 0 ∴ x ∈ {-3, 4}
x2-5x = 24 x2-5x-24 = 0 (x-8)(x+3) = 0 x = 8 or x -3
x2+4x-9 = 5x+3 x2+4x-5x-9-3 = 0 x2-x-12 = 0 (x+3)(x-4) = 0 x = -3 or x = 4
X2 - 5X = 24 X2 - 5X - 24 = 0 (X - 8)(X + 3) X = 8 ----------------and X = - 3 --------------
If you mean: x2+5x+6 = 0 then the solutions are x = -3 and x = -2
the question is to solve (x^2-5x+5)^(x^2-36)=1
x2 + 5x - 120 can not be factored.
You don't factor an equation. You factor an expression.So we'll forget about the "equals 0" for a second, while we factor "x2 plus 5x plus 6".x2 + 5x + 6 = (x + 3) (x + 2).Now, if you want this expression to be equal to zero, that can only happen when 'x' is -2 or -3.
x2 + 5x + 6 = (x + 2)(x + 3)