x2+8x+9 = -7 x2+8x+9+7 = 0 x2+8x+16 = 0 (x+4)(x+4) = 0 Therefore: x = -4 and also x = -4 (they both have equal roots)
2x2 + x2 = 3x2
x2 +6x = 27 x2 + 6x - 27 = 0 x2 + 9x - 3x - 27 = 0 x(x + 9) - 3(x + 9) =0 (x - 3)(x + 9) = 0 So x can equal 3 or -9.
x2 + x2 = 2x2 2x^2
x2-6x+9 = (x-3)(x-3) when factorised.
x2+8x+9 = -7 x2+8x+9+7 = 0 x2+8x+16 = 0 (x+4)(x+4) = 0 Therefore: x = -4 and also x = -4 (they both have equal roots)
It is: (-4)^2 +(-4) -9 = 3 when x is equal to -4
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}
It equals x2 - 34
2x2 + x2 = 3x2
x2 +6x = 27 x2 + 6x - 27 = 0 x2 + 9x - 3x - 27 = 0 x(x + 9) - 3(x + 9) =0 (x - 3)(x + 9) = 0 So x can equal 3 or -9.
x2+24x+144 = 9 x2+24x+144-9 = 0 x2+24x+135 = 0 (x+9)(x+15) = 0 x = -9 or x = -15
x2 + 9 = 10x x2 - 10x + 9 = 0. (x - 9)(x - 1) = 0. Therefore, x = 1 or 9.
It is the circular disc with centre at the origin and radius = 3 units.
x2 + 12x + 27 = (x + 9)(x + 3)
x2 + x2 = 2x2 2x^2
x2-6x+9 = (x-3)(x-3) when factorised.