-10
x2 + 6x + 9 = 81 x2 + 6x = 72 x2 + 6x - 72 = 0 (x+12)(x-6) = 0 x= -12, 6 (two solutions)
(x - 12)(x + 12)
x2 + 6x + 12 = 0 x2 + 6x + 9 = -3 (x + 3)2 = -3 x + 3 = ± √-3 x = -3 ± i√3
x2 - 4x = 8 x2 - 4x + 4 = 8 + 4 (x - 2)2 = 12 x - 2 = +&- sq. root of 12 x = 2 +&- 2(sq. root of 3)
x2 - 10x - 24= x2 - 12x + 2x - 24= x(x - 12) + 2(x - 12)= (x - 12)(x + 2)
14
If: x2+x = 12 Then: x2+x-12 = 0 And using the quadratic formula: x = -4 or x = 3
(x2-x-12)/(x-4) = (x+3)
x2 -11x-12= (x-12)(x+1)
14
No. x2 -4x - 12 = (x + 2)(x - 6)
x2 + 13x + 12 = (x + 1)(x + 12)
The anti-derivative of X2 plus X is the same as the anti-derivative of X2 plus the anti-derivative of X. The anti derivative of X2 is X3/3 plus an integration constant C1 The anti derivative of X is X2/2 plus an integration constant C2 So the anti-derivative of X2+X is (X3/3)+(X2/2)+C1+C2 The constants can be combined and the fraction can combined by using a common denominator leaving (2X3/6)+(3X2/6)+C X2/6 can be factored out leaving (X2/6)(2X+3)+C Hope that helps
-10
x2 - x - 12 = 0 ∴ (x + 3)(x - 4) = 0 ∴ x ∈ {-3, 4}
x2 + x - 12 = (x + 4)(x - 3)