Can You Solve This Challenging Polynomial Equation?

[anemone]

Solve $(x^2-9x+22)^2-10(x^2-9x+22)=x-22$.

 

[topsquark]

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

 

[anemone]

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

Spoiler

witty and humorous...

(Tongueout)

 

[anemone]

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)

 

[MarkFL]

My solution:

Spoiler

Let's arrange the equation as:

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

Expanding and collecting like terms yields:

\(\displaystyle x^4-18x^3+116x^2-316x+308=0\)

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

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

Expand the right side:

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

Equating coefficients, we obtain the system:

\(\displaystyle a+c=-18\)

\(\displaystyle ac+b+d=116\)

\(\displaystyle ad+bc=-316\)

\(\displaystyle bd=308\)

Solving this system, we obtain:

\(\displaystyle a=-10,\,b=22,\,c=-8,\,d=14\)

Hence:

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

Application of the quadratic formula yields:

\(\displaystyle x=4\pm\sqrt{2},\,5\pm\sqrt{3}\)

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

\(\displaystyle u=x^2-9x+22\)

And the original then becomes:

\(\displaystyle u^2-9u+22=x\)

Adding the two equations, we obtain:

\(\displaystyle u^2-8u+22=x^2-8x+22\)

And this may be reduced to:

\(\displaystyle (u-x)(u+x-8)=0\)

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

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

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

And the roots follow as above:

\(\displaystyle x=4\pm\sqrt{2},\,5\pm\sqrt{3}\) :D

 

[Albert1]

Solve $(x^2-9x+22)^2-9(x^2-9x+22)=x-22---(1)$.

Spoiler

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
$x=4\pm \sqrt 2,\,\ or\,\, x=5\pm \sqrt 3$

 

[anemone]

My solution:

Spoiler

Let's arrange the equation as:

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

Expanding and collecting like terms yields:

\(\displaystyle x^4-18x^3+116x^2-316x+308=0\)

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

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

Expand the right side:

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

Equating coefficients, we obtain the system:

\(\displaystyle a+c=-18\)

\(\displaystyle ac+b+d=116\)

\(\displaystyle ad+bc=-316\)

\(\displaystyle bd=308\)

Solving this system, we obtain:

\(\displaystyle a=-10,\,b=22,\,c=-8,\,d=14\)

Hence:

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

Application of the quadratic formula yields:

\(\displaystyle x=4\pm\sqrt{2},\,5\pm\sqrt{3}\)

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

\(\displaystyle u=x^2-9x+22\)

And the original then becomes:

\(\displaystyle u^2-9u+22=x\)

Adding the two equations, we obtain:

\(\displaystyle u^2-8u+22=x^2-8x+22\)

And this may be reduced to:

\(\displaystyle (u-x)(u+x-8)=0\)

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

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

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

And the roots follow as above:

\(\displaystyle x=4\pm\sqrt{2},\,5\pm\sqrt{3}\) :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:

Spoiler

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!:)

 

[Albert1]

my solution is the same as MarkFL's second
part solution ,sorry I did not know he post it first

 

[MarkFL]

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)

 

[MarkFL]

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