(x^2-3x+2)^2-3(x^2-3x+2)-2-x=0 (Moldova 2000 Grade 9 P5)

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(x^2-3x+2)^2-3(x^2-3x+2)-2-x=0 (Moldova 2000 Grade 9 P5)

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#1 h Apr 26, 2021, 12:42 AM

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Solve in real numbers the equation
$\left(x^2-3x-2\right)^2-3\left(x^2-3x-2\right)-2-x=0.$

This post has been edited 1 time. Last edited by jasperE3, Apr 26, 2021, 1:01 PM
Reason: x^2-3x-2 instead of x^2-3x+2

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Mathematician1010

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#2 h Apr 26, 2021, 1:24 AM

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Start out by simplifying the left-hand side, and you get
$x^4-6x^3+10x^2-4x-4=0$
After that, I'd recommend using the rational root theorem, and if there are any rational solutions, you can divide out the factor. I haven't done this for this problem yet, but if there are two rational solutions you can reduce down to a quadratic and find any other solutions with the quadratic formula.

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#3 h Apr 26, 2021, 3:47 AM

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RRT gives no rational roots, you'd probably have to factor it into two quadratics. Any solutions without undetermined coefficients?

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#4 h Apr 26, 2021, 4:02 AM

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progress

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#5 h Apr 26, 2021, 4:11 AM • 3 Y

Y by Mango247, Mango247, Mango247

yofro (via PM) wrote:

What happens if x is a root of $x^2-4x-2$ ?

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#6 h Apr 26, 2021, 4:30 AM

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Post #5 by jasperE3

@jasperE3 ok

soo... some more progress

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#7 h Apr 26, 2021, 4:37 AM

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jasperE3 wrote:

yofro (via PM) wrote:

What happens if x is a root of $x^2-4x-2$ ?

That's pretty cool
Click to reveal hidden text

The $x^2-4x-2$ is kind of intuitive because clearly setting $x^2-3x-2=x$ gives a solution to the above equation. We still need the other quadratic factor for 4 roots, but I'm not seeing something that is intuitive. Of course, it's pretty easy now to just factor $x^2-4x-2$ from the equation in #2 and use MUD, but I was wondering if there was a more intuitive approach on that other factor.

@below you really should be thanking yofro for these amazing hints. Do you have a reason to believe the other factor does not have real roots?

This post has been edited 2 times. Last edited by natmath, Apr 26, 2021, 4:55 AM

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#8 h Apr 26, 2021, 4:42 AM • 3 Y

Y by Mango247, Mango247, Mango247

Post #7 by natmath

@natmath whoa that's awesome

solutions to $x^2-4x-2$ are $x=\frac{4\pm\sqrt{16+8}}{2}=\frac{4\pm2\sqrt6}{2}=2\pm\sqrt6.$ Those are real! 2 solutions to the equation already!

If I'm not mistaken, these are the only real solutions and this problem is solved....?

This post has been edited 1 time. Last edited by Lamboreghini, Apr 26, 2021, 4:43 AM

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#9 h Apr 26, 2021, 5:25 AM

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uh I think you messed up? You're confusing $x^2-4x+2$ and $x^2-4x-2$ ? Is there a typo in the problem?

If there's a typo in the problem and it is $x^2-3x-2$ instead of $x^2-3x+2$ we can use poly division and get our other answers to be $1\pm \sqrt 5$

and of course using the other roots, we can let $x$ be a root of $x^2-2x-4$ , then $x^2-3x-2=-x+2$ , and $(-x+2)^2-3(-x+2)-2-x=x^2-4x+4+3x-6-2-x=x^2-2x-4=0$

This post has been edited 4 times. Last edited by asbodke, Apr 26, 2021, 5:33 AM

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Mathematician1010

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#10 h Apr 26, 2021, 12:27 PM

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@above where are you getting $x^2-2x-4$ from?

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#11 h Apr 26, 2021, 1:01 PM

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Very sorry. Yes, the problem was incorrect. It has been edited now.

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#12 h Apr 26, 2021, 3:18 PM • 1 Y

Y by yofro

Yofro gave me a hint on $x^2-2x-4$ . It's not as direct as the first one, but I can try to explain it.

If $f(x)=x^2-3x-2$ , then we want to find the solutions to
$f(f(x))=x$
Of course, if there existed an $r$ s.t. $f(r)=r$ , then this would clearly be a solution to our equation. That $r$ must be either of the roots to
$x^2-3x-2=x$ $x^2-4x-2=0$
Let's say for some constant $k$ there exists an $r$ s.t. $f(r)=k-r$ . Now let's say $k-r$ also satisfies the equation $f(x)=k-x$ (i.e. $f(k-r)=k-(k-r)=r$ ). Then this $r$ would satisfy the original equation.
However, finding this root is not as immediate. We want $r$ and $k-r$ to be the solutions of
$f(x)=k-x$ $x^2-3x-2=k-x$ $x^2-2x-(2+k)=0$ By viete's, the sum of roots is
$r+k-r=2$ $k=2$ So the solutions to
$x^2-2x-4=0$ also satisfy our original equation.

Also I think we know that there are no other solutions aside from those in the form $f(x)=x$ and $f(x)=k-x$ because $x,k-x$ are the only polynomial functions that satisfy $f(f(x))=x$ .

This post has been edited 1 time. Last edited by natmath, Apr 26, 2021, 3:29 PM

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#13 h Apr 26, 2021, 4:07 PM

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Take one x in other side and set y= $x^2-3x$ and solve quadratic take one of its root equate it to x then you get a good answer

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