Solving the equation $(x^2-3x+1)^2=4x^2-12x+9$ I need to solve the following equation:
$$(x^2-3x+1)^2=4x^2-12x+9.$$
I think I need to bring everything to one side but I don't know anything else.
 A: HINT:
$$(x^2-3x+1)^2=(2x-3)^2$$
Now use $a^2-b^2=(a+b)(a-b)$
A: Given equation : $(x^2-3x+1)^2=4x^2-12x+9$  ....(1)
To solve this equation we need the identity,
$a^2-2ab+b^2=(a-b)^2$
on observing the R.H.S. of the equation (1) we can write $4x^2-12x+9$ as
$(2x)^2-2\times2x\times3+3^2$ 
Now from the above identity, it can be written as $(2x-3)^2$
Thus, equation (1) becomes $(x^2-3x+1)^2=(2x-3)^2$
$\implies$ $(x^2-3x+1)^2-(2x-3)^2=0$
$\implies$ $(x^2-3x+1+2x-3)(x^2-3x+1-2x+3)=0$ $\;\;\;\;$$\because\,a^2-b^2=(a+b)(a-b)$
$\implies$ $(x^2-x-2)(x^2-5x+4)=0$
$\implies$ $(x^2-2x+x-2)(x^2-x-4x+4)=0$
$\implies$ $\{x(x-2)+1(x-2)\}\{x(x-1)-4(x-1)\}=0$
$\implies$ $\{(x-2)(x+1)\}\{(x-1)(x-4)\}=0$
$\implies$ $(x-2)(x+1)(x-1)(x-4)=0$
Thus, x = 2, -1, 1, 4 
this is the solution of given equation.
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Answer = $$(x^2-3x+1)^2=4x^2-12x+9$$ $$(x^2-3x+1)^2=(2x-3)^2$$ 
$$(x^2-3x+1)^2-(2x-3)^2=0$$ $$x^2-5x+4=0$$ $$x^2-x-2=0$$ 
$$x=4$$
$$x=1$$
$$x=2$$
$$x=-1$$
A: $$(x^2-3x+1)^2=4x^2-12x+9$$
$$(x^2-3x+1)^2=4(x^2-3x+1)+5$$
$$(x^2-3x+1)^2-4(x^2-3x+1)-5=0$$
now let $u=(x^2-3x+1)^2$
so
$$u^2-4u-5=0$$
$$(u+1)(u-5)=0$$
then complete the solution
A: (x$^2$ - 3x + 1)$^2$ = (2x - 3)$^2$
(x$^2$ - 3x + 1)$^2$ - (2x - 3)$^2$ = 0
(x$^2$-3x+1-2x+3)(x$^2$-3x+1+2x-3) = 0
(x$^2$-5x+4)(x$^2$-x-2) = 0
(x-1)(x-4)(x+1)(x-2) = 0
Therefore, solution is x = 1, x = 4, x = -1 or x = 2
A: First you need to factor before bring the terms into one side. $$(x^2-3x+1)^2=4x^2-12x+9$$ $$(x^2-3x+1)^2=(2x-3)^2$$ Now you subtract the right hand side, where it becomes $$(x^2-3x+1)^2-(2x-3)^2=0$$ You can factor by using differences of squares where $a^2-b^2=(a+b)(a-b)$. 
Now finish this off on your own:)
If you really want the full solution just mouse over this part

 $$(x^2-3x+1)^2-(2x-3)^2=0 $$ $$(x^2-x-2)(x^2-5x+4)=0$$ $$(x-2)(x+1)(x-4)(x-1)=0$$ $$x=-1,1,2,4$$

