(x + 3)(x + 4)
x2 = -11x - 10 x2 + 11x + 10 = 0 (x + 1)(x + 10) = 0
(x-8)(x+8) = x2-64
x2 + 11x + 18 (x + 9)(x + 2) CHECK: x2 + 9x + 2x + 18 x2 + 11x + 18 SET EACH EQUAL TO ZERO: x + 9 = 0 x = -9 x + 2 = 0 x = -2 NOW YOU ARE DONE: Solution set: {-9, -2}
x2-5x = 0 x(x-5) = 0 x = 0 or x = 5
(x + 3)(x + 4)
x = 0
(x + 6)(x - 6)
x2=9x-20 x2-9x+20=0 Factor: (x-4)(x-5)=0 x={4,5}
x2+7x+12=0 x2+7x=-12 x+Sqaure Root of 7x= 2(SQ Root Symbol)3
x3=x2 x3-x2=0 Factor out x2 x2(x - 1) = 0 If x2=0 then x=0. If x - 1 = 0 then x = 1. Therefore the number is 0 or 1.
x2 - 7x = 60 x2 - 7x - 60 = 0 (x - 12) (x + 5) = 0 x = 12 or x = -5
x2-4x-21 = 0 => x = -3 or x = 7 x2-3x-18 = 0 => x = -3 or x = 6
x2 - 9x = -8 ∴ x2 - 9x + 8 = 0 ∴ (x - 1)(x - 8) = 0 You can then solve it as well: ∴ x ∈ {1, 8}
(x - 3)(x - 2)
x2+7x+12=0usually this means to factor the equation.factored this equation would be...(x+3)(x+4)
If: x2+x = 12 Then: x2+x-12 = 0 And using the quadratic formula: x = -4 or x = 3