Solve the equation : ##x^4+12x^3+2x^2+25=0##

[chwala]

Homework Statement

Solve the equation $x^4+12x^3+2x^3+25=0$

Relevant Equations

quartic equation

Wolfram gave the solution and a hint: i want to understand the hands on approach steps...

In my approach (following Wolfram's equation) i have,

$(x-3)^2(2+12(x-3)+(x-3)^2=-25$

$(x-3)^2((x+3)^2-33)=-25$

$(x-3)\sqrt{((x+3)^2-33)}=-5i$

...

 

[anuttarasammyak]

Wolfram says x^3 term vanish by the substitution.

 

[BvU]

$(x-3)^4 = 81 - 108 x + 54 x^2 - 12 x^3 + x^4$

$\$

 

[Steve4Physics]

Homework Statement: Solve the equation $x^4+12x^3+2x^3+25=0$
Relevant Equations: quartic equation

Wolfram gave the solution and a hint: i want to understand the hands on approach steps...

You seem to have typo’s in the Title and the Homework Statement: presumably $2x^3$ should be $2x^2$.

I’m not sure what you are asking. By using the substitution u=x-3 $u=x+3$ you can convert the equation into a ‘depressed quartic’ from form ($u^4+au^2+bu+c=0$; note there is no $u^3$ term). I guess that’s the gist of what Wolfram is suggesting. You then solve the depressed cubic quartic.

The technique is described in the Wiki article on quartics here: https://en.wikipedia.org/wiki/Quartic_equation

That's my guess.

Edits: as indicated by the strike-throughs!

 

[SammyS]

Homework Statement: Solve the equation $x^4+12x^3+2x^3+25=0$
Relevant Equations: quartic equation

Wolfram gave the solution and a hint: i want to understand the hands on approach steps...

In my approach (following Wolfram's equation) i have,

$(x-3)^2(2+12(x-3)+(x-3)^2=-25$
...

Here's a snip from your pdf.

So, Steve is correct. You have a typo.

The problem that you gave to Wolfram Alpha Is to solve the equation:

$\displaystyle \quad x^4+12x^3+2x^2+25=0$

Added in Edit: (Just about the same time Steve posted the following Post.)
The hint you, @chwala , were given could be more clearly stated as: Let $\displaystyle\ u=x+3$ . i.e. $\displaystyle \ x=u-3$ .

Look at the binomial expansions of $\displaystyle \ (u-3)^4\$ and $\displaystyle \ 12(u-3)^3\$.

 

[Steve4Physics]

So, Steve is correct. You have a typo.

Thanks. In fact I'd messed-up so would like to add this...

I don't like the Wolfram instructions/notation to substitute '$x=x+3$' which seems ambiguous/confusing.

I think Wolfram is saying: use $u=x+3$, i.e. $x=u-3$, so that the original equation:
$x^4 +12x^3 + 2x^2 + 25 = 0$
becomes:
$(u-3)^4 +12(u-3)^3 + 2(u-3)^2 + 25 = 0$
which, when expanded, gives the more manageable depressed quartic form.

 

[chwala]

You seem to have typo’s in the Title and the Homework Statement: presumably $2x^3$ should be $2x^2$.

I’m not sure what you are asking. By using the substitution u=x-3 $u=x+3$ you can convert the equation into a ‘depressed quartic’ from form ($u^4+au^2+bu+c=0$; note there is no $u^3$ term). I guess that’s the gist of what Wolfram is suggesting. You then solve the depressed cubic quartic.

The technique is described in the Wiki article on quartics here: https://en.wikipedia.org/wiki/Quartic_equation

That's my guess.

Edits: as indicated by the strike-throughs!

Let me fix that to mitigate any confusion for those following the thread...

 

[topsquark]

Let me fix that to mitigate any confusion for those following through the thread...

What methods for solving quartics have you studied?

-Dan

 

[chwala]

Nice we then have,

$(u-3)^4+12(u-3)^3+2(u-3)^2+25=0$

$(u^4-12u^3+54u^2-108u+81)+12(u^3-9u^2+27u-27)+2u^2-12u+18+25=0$

$u^4-52u^2+204u-200=0$

$u^4=52u^2-204u+200$

...i am trying to follow this:

in my working i have,

$j=\dfrac{-200}{k}$

$k=\dfrac{1}{2}h^2-26-102h$

implying that

$j=\dfrac{1}{2}h^2-26-102h$

using $\dfrac{q^2}{h^2} +4r$

i do not seem to see how $h^4$ is eliminated to realize the cubic equation.
... This approach https://en.wikipedia.org/wiki/Quartic_equation seems to be ideal i will need to check and follow on it later. ...

 

[chwala]

I managed to use the approach of Factoring Quartic into a product of two quadratics. This approach is straightforward.

$au^4+bu^3+cu^2+du+e = (u^2+pu+q)(u^2+ru+s)$
...
I then had the cubic equation,

$P^3-104P^2+3504P-41616=0$

Where $P=p^2$

and

$0=p+r$
$\dfrac{d}{p}=s-q$
$d=ps+qr$

$P=52.0199, ⇒p=7.212$

thus,

$u^4-52u^2+204u-200= (u^2+7.212u-14.15)(u^2-7.212u+14.15)=0$

 

[anuttarasammyak]

Good job! Why don't you fiish it to show us two real solutions and two conjugate complex solutions which Wolfram suggests ?

 

[chwala]

Good job! Why don't you fiish it to show us two real solutions and two conjugate complex solutions which Wolfram suggests ?

Most definitely; having a coffee break now...laters.

 

[chwala]

Good job! Why don't you fiish it to show us two real solutions and two conjugate complex solutions which Wolfram suggests ?

using quadratic formula on

$u^2+7.212u-14.15=0$,

$u_1≈ -8.8168$

$u_2≈1.6048$

$u_3 ,u_4≈\left[ \dfrac{7.212±\sqrt{-4.6}}{2}\right]≈\left[3.606±\dfrac{2.14476i}{2}\right]≈3.606±1.07238i$

The solutions to our original problem are therefore,

$x_1≈-11.8168$ and $x_2≈-1.3952$

The textbook page below prompted me to seek for a method that would solve the quartic equation. Cheers guys.

Bingo!

 

[topsquark]

The textbook page below prompted me to seek for a method that would solve the quartic equation. Cheers guys.

View attachment 335280

Bingo!

I disagree with your source. You do not need to use any form of computational aid. There are ways to exactly solve quartic equations. Ferrari's method, for example, always works.

However, solving a general quartic by hand can be a real mess. (I admit that I've never actually managed to get the correct answer using Ferrari's method, except for the example problem I saw that was fiddled with to make everything easy.) I agree that you would probably find a computational aid to be quite useful, especially if all you need are approximate solutions.

-Dan

 

[chwala]

I disagree with your source. You do not need to use any form of computational aid. There are ways to exactly solve quartic equations. Ferrari's method, for example, always works.

However, solving a general quartic by hand can be a real mess. (I admit that I've never actually managed to get the correct answer using Ferrari's method, except for the example problem I saw that was fiddled with to make everything easy.) I agree that you would probably find a computational aid to be quite useful, especially if all you need are approximate solutions.

-Dan

Ferrari's method I admit took me round in circles...one has to be keen and accurate with the values...spent time on it without making any meaningful progress...

 

[topsquark]

Ferrari's method I admit took me round in circles...one has to be keen and accurate with the values...spent time on it without making any meaningful progress...

Well, that's because the exact form of the roots in this case are horrible. (It also tends to happen when the form of the roots are nice. It's just an ugly business.) All I am saying is that it can be done. I rarely even solve cubics exactly, which isn't too hard with a little practice, but you commonly get some really awkward expressions for integers, so I usually just use a calculator. (And I don't often need an exact expression anyway, as I'm a Physicist and not a Mathematician.)

-Dan

 

[anuttarasammyak]

The textbook page below prompted me to seek for a method that would solve the quartic equation. Cheers guys.

Bingo!

The textbook page below prompted me to seek for a method that would solve the quartic equation. Cheers guys.

As the graph shows
$$y=x^4+12x^3+2x^2+25$$
$$y(0)=25 > 0$$
$$y'(0)=y"(0)＝0$$
x=0 is an inflection point. Usually x^4 and smaller order formula show double well shape but this one has only one well. We know the equation has two real solutions of minus sign.
$$y(-1)=16 >0$$
$$y(-2)=16-96+8+25=-47<0$$
We know a solution lies in (-2,-1).
$$y(-12)=2*144+25=313 >0$$ which comes back to small plus.
We may suspect that another one lies in (-12,-11).

 

[SammyS]

As the graph shows
$$y=x^4+12x^3+2x^2+25$$ $$y(0)=25 > 0$$ $$y'(0)=y"(0)＝0$$ x=0 is an inflection point. Usually x^4 and smaller order formula show double well shape but this one has only one well.

$\displaystyle y''(0)=4\ne0$

There is an inflection point at $\displaystyle x=-3+\sqrt{\dfrac{26}{3}} \approx -0.05608$ .

There is a double well shape, but the well at $\displaystyle x=0$ is very shallow. The relative maximum at $\displaystyle x=\dfrac{-9+\sqrt{77}}{2} \approx -0.11252$ is approximately $0.008387$ higher than the relative minimum of $25$ at $x=0$ .

 

[chwala]

$\displaystyle y''(0)=4\ne0$

There is an inflection point at $\displaystyle x=-3+\sqrt{\dfrac{26}{3}} \approx -0.05608$ .

There is a double well shape, but the well at $\displaystyle x=0$ is very shallow. The relative maximum at $\displaystyle x=\dfrac{-9+\sqrt{77}}{2} \approx -0.11252$ is approximately $0.008387$ higher than the relative minimum of $25$ at $x=0$ .

@SammyS wow! You see deeper...that's new will check on the graph keenly later...

 

[chwala]

@SammyS wow! You see deeper...that's new will check on the graph keenly later...

We do have two inflection points, that is at

$(x_1, y_1)= (-5.944,-1176.115)$ & $(x_2,y_2)=( -0.0561,25.00)$.

 

[anuttarasammyak]

I also check it for my study
$$y(u)=u^4-52u^2+204u-200$$
where
$$u=x+3$$
$$y^"=12u^2-104$$
In order y"=0
$$x=-3\pm\sqrt{\frac{26}{3}}$$