MHB Can You Solve This Challenging Polynomial Equation?

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The discussion revolves around solving the polynomial equation $(x^2-9x+22)^2-10(x^2-9x+22)=x-22$, later corrected to $(x^2-9x+22)^2-9(x^2-9x+22)=x-22$. Participants express their challenges and excitement in tackling the problem, highlighting various methods to solve polynomial equations. Several users, including Dan, MarkFL, and Albert, share their solutions and acknowledge each other's contributions. The conversation emphasizes collaboration and the enjoyment of problem-solving within the mathematical community.
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Solve $(x^2-9x+22)^2-10(x^2-9x+22)=x-22$.
 
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Well, I couldn't do it. On the other hand I didn't realize how many ways I know how to solve a polynomial equation, so it's been a good problem for me. (I drew the line when it came to solving a general quartic. Even I'm not that crazy! (Sun) )

-Dan
 
topsquark said:
Well, I couldn't do it. On the other hand I didn't realize how many ways I know how to solve a polynomial equation, so it's been a good problem for me. (I drew the line when it came to solving a general quartic. Even I'm not that crazy! (Sun) )

-Dan

I was so excited when this new post came into my today challenge problem! And you sure are
witty and humorous...
(Tongueout)
 
anemone said:
Solve $(x^2-9x+22)^2-10(x^2-9x+22)=x-22$.

Hi MHB,

I only realized by now that there is a typo, an error in the coefficient of 10 that is attached to the term $x^2-9x+22$ above, the problem should read:

Solve $(x^2-9x+22)^2-9(x^2-9x+22)=x-22$.

I am terribly sorry that I did that.(Tmi)(Doh)
 
My solution:

Let's arrange the equation as:

$$\left(x^2-9x+22\right)^2-9\left(x^2-9x+22\right)-x+22=0$$

Expanding and collecting like terms yields:

$$x^4-18x^3+116x^2-316x+308=0$$

Now, let's assume we may factor as follows:

$$x^4-18x^3+116x^2-316x+308=\left(x^2+ax+b\right)\left(x^2+cx+d\right)$$

Expand the right side:

$$x^4-18x^3+116x^2-316x+308=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd$$

Equating coefficients, we obtain the system:

$$a+c=-18$$

$$ac+b+d=116$$

$$ad+bc=-316$$

$$bd=308$$

Solving this system, we obtain:

$$a=-10,\,b=22,\,c=-8,\,d=14$$

Hence:

$$x^4-18x^3+116x^2-316x+308=\left(x^2-10x+22\right)\left(x^2-8x+14\right)$$

Application of the quadratic formula yields:

$$x=4\pm\sqrt{2},\,5\pm\sqrt{3}$$

Now, that is a mundane way to solve such a fun equation. Let's look at the substitution:

$$u=x^2-9x+22$$

And the original then becomes:

$$u^2-9u+22=x$$

Adding the two equations, we obtain:

$$u^2-8u+22=x^2-8x+22$$

And this may be reduced to:

$$(u-x)(u+x-8)=0$$

Back-substituting for $u$, we then get:

$$(x^2-9x+22-x)(x^2-9x+22+x-8)=0$$

$$(x^2-10x+22)(x^2-8x+14)=0$$

And the roots follow as above:

$$x=4\pm\sqrt{2},\,5\pm\sqrt{3}$$ :D
 
anemone said:
Solve $(x^2-9x+22)^2-9(x^2-9x+22)=x-22---(1)$.
let $x^2-9x+22=y---(1)$
(1) becomes :$y^2-9y+22=x---(2)$
(1)-(2) we have :$(x-y)(x+y-8)=0--(3)$
$\therefore \,\,x=y---(4),\,\, or \,\,x+y-8=0--(5)$
put (4)(5) to (1) or (2) we get
$x=4\pm \sqrt 2,\,\ or\,\, x=5\pm \sqrt 3$
 
MarkFL said:
My solution:

Let's arrange the equation as:

$$\left(x^2-9x+22\right)^2-9\left(x^2-9x+22\right)-x+22=0$$

Expanding and collecting like terms yields:

$$x^4-18x^3+116x^2-316x+308=0$$

Now, let's assume we may factor as follows:

$$x^4-18x^3+116x^2-316x+308=\left(x^2+ax+b\right)\left(x^2+cx+d\right)$$

Expand the right side:

$$x^4-18x^3+116x^2-316x+308=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd$$

Equating coefficients, we obtain the system:

$$a+c=-18$$

$$ac+b+d=116$$

$$ad+bc=-316$$

$$bd=308$$

Solving this system, we obtain:

$$a=-10,\,b=22,\,c=-8,\,d=14$$

Hence:

$$x^4-18x^3+116x^2-316x+308=\left(x^2-10x+22\right)\left(x^2-8x+14\right)$$

Application of the quadratic formula yields:

$$x=4\pm\sqrt{2},\,5\pm\sqrt{3}$$

Now, that is a mundane way to solve such a fun equation. Let's look at the substitution:

$$u=x^2-9x+22$$

And the original then becomes:

$$u^2-9u+22=x$$

Adding the two equations, we obtain:

$$u^2-8u+22=x^2-8x+22$$

And this may be reduced to:

$$(u-x)(u+x-8)=0$$

Back-substituting for $u$, we then get:

$$(x^2-9x+22-x)(x^2-9x+22+x-8)=0$$

$$(x^2-10x+22)(x^2-8x+14)=0$$

And the roots follow as above:

$$x=4\pm\sqrt{2},\,5\pm\sqrt{3}$$ :D

Well done, MarkFL! (Clapping)

If I were you, I would only show to the world of the second very smart solution, at least that could, hopefully, to impress some, hehehe...(Happy)

Edit:

Albert said:
let $x^2-9x+22=y---(1)$
(1) becomes :$y^2-9y+22=x---(2)$
(1)-(2) we have :$(x-y)(x+y-8)=0--(3)$
$therefore \,\,x=y---(4),\,\, or \,\,x+y-8=0--(5)$
put (4)(5) to (1) or (2) we get
$x=4\pm \sqrt 2,\,\ or\,\, x=5\pm \sqrt 3$

Bravo, Albert, and thanks for participating!:)
 
my solution is the same as MarkFL's second
part solution ,sorry I did not know he post it first
 
Last edited:
anemone said:
Well done, MarkFL! (Clapping)

If I were you, I would only show to the world of the second very smart solution, at least that could, hopefully, to impress some, hehehe...(Happy)

Well, I typed up the first method, but I knew there had to be a more clever way to solve it, so I looked until I found it, and then just added what I had found. Everyone already knows how awesome I am, so there's no need to try to impress...(Smirk)
 
  • #10
Albert said:
my solution is the same as MarkFL's second
part solution ,sorry I did not know he post it first

We were working on them at the same time, and we just happened to come up with the same method. :D
 

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