Answer:
x = 1, x = 2, x = 3
Step-by-step explanation:
Hello, please consider the following.
[tex](x^2-4x)^2+7x^2-28x+12=0 \\\\(x(x-4))^2+7x(x-4)+12=0\\\\\text{Let's note }X = x(x-4)\text{ the equation becomes}\\\\X^2+7X+12=0\\\\\text{The sum of the zeroes is -7=-3+(-4) and the product is 12=(-4)*(-3).}\\\\\text{Wo we can factorise.}\\\\X^2+7X+12=X^2+4X+3X+12=X(X+4)+3(X+4)=(X+4)(X+3)=0\\\\X=-4 \ \ or \ \ X=-3[/tex]
So we need to solve:
[tex]x(x-4)=-4 \ \ or \ \ x(x-4)=-3\\\\<=> x^2-4x-4=(x-2)^2=0 \ \ or \ \ x^2-4x+3=(x-1)(x-3)=0[/tex]
So, the solutions are x = 2, x = 1, x = 3
Thank you
