Question

Hi i need help asap my parent taught me how to do this equation (x^2-4x)^2+7x^2-28x+12=0 but i still dont understand I really need help understanding this i know the answer i just need an explanation please help me

Answer 1

Answer:

x = 1,  x = 2,  x = 3

Step-by-step explanation:

Hello, please consider the following.

$(x^2-4x)^2+7x^2-28x+12=0 \\\\(x(x-4))^2+7x(x-4)+12=0\\\\\text{Let's note }X = x(x-4)\text{ the equation becomes}\\\\X^2+7X+12=0\\\\\text{The sum of the zeroes is -7=-3+(-4) and the product is 12=(-4)*(-3).}\\\\\text{Wo we can factorise.}\\\\X^2+7X+12=X^2+4X+3X+12=X(X+4)+3(X+4)=(X+4)(X+3)=0\\\\X=-4 \ \ or \ \ X=-3$

So we need to solve:

$x(x-4)=-4 \ \ or \ \ x(x-4)=-3\\\\ x^2-4x-4=(x-2)^2=0 \ \ or \ \ x^2-4x+3=(x-1)(x-3)=0$

So, the solutions are x = 2, x = 1, x = 3

Thank you