x2 + 5x + 3 = 9 x2 + 5x - 6 = 0 (x + 6)(x - 1) = 0 x = -6 or x = 1
To solve for x: x2 = 5x + 2 x2 - 5x = 2 x2 - 5x + (5/2)2 = 2 + (5/2)2 (x - 5/2)2 = 8/4 + 25/4 x - 5/2 = ± √(33/4) x = (5 ± √33) / 2
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 - 120 can not be factored.
It is, as stated, x2 - 5x + 25/4 = 0
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
To solve for x: x2 = 5x + 2 x2 - 5x = 2 x2 - 5x + (5/2)2 = 2 + (5/2)2 (x - 5/2)2 = 8/4 + 25/4 x - 5/2 = ± √(33/4) x = (5 ± √33) / 2
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
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)
2x + 5x = 7x
x2 + 5x = 60 x2 + 5x + 25/4 = 60 + 25 / 4 (x + 5/2)2 = 265/4 x + 5/2 = (265/4)1/2 x = -5/2 ± √265 / 2 x = -(5 ± √265) / 2