x2+9x+18 find factors of 18 that add up to 9, in this case 6&3 (x+6)(x+3)
x2 + 3x - 18 = 0 (x - 3)(x + 6) = 0 x = 3 or -6
x2+7x=-2x2+6 3x2+7x-6=0 (3x-2)(x+3)=0 3x-2=0 or x+3=0 x=2/3 or x=-3
x^2+3x-18=0 (x+6)(x-3)=0
x5 - 2x4 - 24x3 = 0 x3(x2 - 2x - 24) = 0 x3(x2 + 4x - 6x - 24) = 0 x3(x - 6)(x + 4) = 0 x = {-4, 0, 6}
If: y = x2-7x+8 and y = -x2+9x-6 Then: x2-7x+8 = -x2+9x-6 So: 2x2-16x+14 = 0 => x2-8x+7 = 0 Therefore: x = 1 and x = 7 By substitution: x =1, y = 2 and x = 7, y = 8 Points of intersection: (1, 2) and (7, 8)
x2 + 7x = 2x2 + 6 x2 - 2x2 + 7x = 6 -x2 +7x - 6 = 0 or alternatively x2 - 7x + 6 = 0
9x-53=6/x 9x2-53x=6 9x2-53x-6=0 9x2+x-54x-6=0 x(9x+1)-6(9x+1)=0 (9x+1)x(x-6)=0 9x+1=0 or x-6=0 9x=-1 or x=6 x=-1/9 or x=6
x4 + 3x3 - x2 - 9x - 6 = 0 x4 + x3 + 2x3 + 2x2 - 3x2 - 3x - 6x - 6 = 0 x3(x + 1) + 2x2(x + 1) - 3x(x + 1) - 6(x + 1) = 0 (x + 1)(x3 + 2x2 - 3x - 6) = 0 (x + 1)[x2(x + 2) - 3(x + 2)] = 0 (x + 1)(x + 2)(x2 - 3) = 0 So x + 1 = 0 so that x = -1 or x + 2 = 0 so that x = -2 or x2 - 3 = 0 so that x = +/- sqrt(3)
x2-9x+18 = (x-3)(x-6) when factorised
-x2 = x - 6 x2 + x - 6 = 0 (x - 2)(x + 3) = 0 x ∈ {-3, 2}
x2 - 2x - 5 = 0 x2 - 2x + 1 = 6 (x - 1)2 = 6 x - 1 = ± √6 x = 1 ± √6
(x + 6)(x - 6)
It is a quadratic equation that can be solved by using the quadratic equation formula whereas x = -9.321825 or x = 0.321825 both given to 6 decimal places
x2- 7x + 6 = 0 factor, (x-6)(x-1) = 0 x = 6 x = 1
x2 - x - 6 = (x - 3)(x + 2) = 0; whence, x = 3 or -2.
x2 - 9x + 18 = (x - 3)(x - 6)