# 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.

• can someone show me the full solve just to check my answer im not sure if i got it Commented Sep 27, 2016 at 5:23

## 6 Answers

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

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

• i know how to solve it now. thanks. Commented Sep 27, 2016 at 5:20
• tytytytytytytyty Commented Sep 27, 2016 at 5:32
• You're welcome. Solving the last part by yourself helps you learn though. Commented Sep 27, 2016 at 5:34

HINT:

$$(x^2-3x+1)^2=(2x-3)^2$$

Now use $a^2-b^2=(a+b)(a-b)$

• i know how to solve it now. thanks. Commented Sep 27, 2016 at 5:20

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 $(x^2-3x+1)^2=(2x-3)^2$

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

$\implies$ $(x^2-3x+1+2x-3)(x^2-3x+1-2x+3)=0$ $\;\;\;\;$$\because\,a^2-b^2=(a+b)(a-b) \implies (x^2-x-2)(x^2-5x+4)=0 \implies (x^2-2x+x-2)(x^2-x-4x+4)=0 \implies \{x(x-2)+1(x-2)\}\{x(x-1)-4(x-1)\}=0 \implies \{(x-2)(x+1)\}\{(x-1)(x-4)\}=0 \implies (x-2)(x+1)(x-1)(x-4)=0 Thus, x = 2, -1, 1, 4 this is the solution of given equation. Easy, download Photmath! Answer =$$(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=0x^2-5x+4=0x^2-x-2=0x=4x=1x=2x=-1(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 (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