First, factorise:
x2 + 5x - 6 = (x + 6)(x - 1) = 0; whence, either
x + 6 = 0, or
x - 1 = 0.
The product of two factors equals zero, exactly when one or other of the two factors equals zero.
Thus, the solution must be x = 1 or -6. If x equals either of those two values, then the original equation will be true.
x2 + 5x + 3 = 9 x2 + 5x - 6 = 0 (x + 6)(x - 1) = 0 x = -6 or x = 1
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 - 4 = 0. Then, x2 - 5x + 4 = (x - 4)(x - 1) = 0; whence, x = 1 or 4. Checking, -1 + 5 - 4 = 0; and -16 + 20 - 4 = 0.
x = 0 or x =-5
x2 + 5x + 3 = 9 x2 + 5x - 6 = 0 (x + 6)(x - 1) = 0 x = -6 or x = 1
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 - 4 = 0. Then, x2 - 5x + 4 = (x - 4)(x - 1) = 0; whence, x = 1 or 4. Checking, -1 + 5 - 4 = 0; and -16 + 20 - 4 = 0.
x = 0 or x =-5
x2 - 5x = 0x (x - 5) = 0x = 0andx = 5
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 x2+5x -6 = 0 (x+6)(x-1) = 0 x+ 6 = 0 x = -6 x-1 = 0 x = 1
It has no solutions because the discriminant of the quadratic equation is less than zero.
(x + 7) + (x - 2) x= -7 or 2
x=8
x2 + 5x - 594 = 0 You need two factors of 594 whose difference by 5. They are 22 and 27. So x = 22 or x = -27