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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.

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  • $\begingroup$ can someone show me the full solve just to check my answer im not sure if i got it $\endgroup$ – hurns Sep 27 '16 at 5:23
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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

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

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  • $\begingroup$ i know how to solve it now. thanks. $\endgroup$ – hurns Sep 27 '16 at 5:20
  • $\begingroup$ tytytytytytytyty $\endgroup$ – hurns Sep 27 '16 at 5:32
  • $\begingroup$ You're welcome. Solving the last part by yourself helps you learn though. $\endgroup$ – suomynonA Sep 27 '16 at 5:34
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HINT:

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

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

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  • $\begingroup$ i know how to solve it now. thanks. $\endgroup$ – hurns Sep 27 '16 at 5:20
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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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Easy, download Photmath!
Answer = $$(x^2-3x+1)^2=4x^2-12x+9$$ $$(x^2-3x+1)^2=(2x-3)^2$$ $$(x^2-3x+1)^2-(2x-3)^2=0$$ $$x^2-5x+4=0$$ $$x^2-x-2=0$$ $$x=4$$ $$x=1$$ $$x=2$$ $$x=-1$$

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$$(x^2-3x+1)^2=4x^2-12x+9$$ $$(x^2-3x+1)^2=4(x^2-3x+1)+5$$ $$(x^2-3x+1)^2-4(x^2-3x+1)-5=0$$ now let $u=(x^2-3x+1)^2$ so $$u^2-4u-5=0$$ $$(u+1)(u-5)=0$$ then complete the solution

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(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

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