Question

Find the real/and or imaginary solutions by factoring:

Answer 1

Answer:

1) x = 9, 3

2)$x = 0, 3+\sqrt{3} , 3+\sqrt{3}$

Step-by-step explanation:

1) $x^2-12x=-27\\x^2-12x+27 = 0$

After factoring:

$(x-9)(x-3) = 0\\x = 9, 3$

2) $x^3-6x^2+6x=0\\x(x^2-6x+6)=0\\x(x-3-\sqrt{3})(x-3+\sqrt{3})$

$x = 0, 3+\sqrt{3} , 3+\sqrt{3}$