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Solving a quartic equation

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anemone

MHB POTW Director
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Feb 14, 2012
3,753
Hi MHB,

I have tried my very best to solve for $x$ in this problem, a quartic with a very large coefficient of $x^2$ but a small coefficient of $x^4$ and $x$ ,but ended up failing miserably...

I began to think it is a function that has no real roots because it just makes me feel better to think of it that way! But seriously though, I think it's a doable problem, but it's just that my ability to solve it is limited.

Could anyone show me the correct approach to crack it?

Thanks in advance.

Problem:

Find all real roots of the quartic $x^4-(2k + 1)x^2 -x + k^2+ k - 1 = 0$ correct to 4 decimal places, where $k= 10^{10}$.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Find all real roots of the quartic $x^4-(2k + 1)x^2 -x + k^2+ k - 1 = 0$ correct to 4 decimal places, where $k= 10^{10}$.
Let $f(x) = x^4-(2k + 1)x^2 -x + k^2+ k - 1$ (where for the moment $k$ is just a large positive number). Then $f(k^{1/2}) = -k^{1/2} - 1 < 0$. But $f$ is positive at $x=0$ and also as $x\to\infty$. So $f$ must have a real root between $0$ and $k^{1/2}$, and another real root greater than $k^{1/2}$.

My next move would be to let $x=k^{1/2}+y$ and get an equation for $y$. I would hope to get some information indicating that $y$ is approximately $\pm k^\alpha$, for some value of $\alpha$ less than $1/2$. But I don't have time to do that just now.
 

topsquark

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MHB Math Helper
Aug 30, 2012
1,140
Hi MHB,

I have tried my very best to solve for $x$ in this problem, a quartic with a very large coefficient of $x^2$ but a small coefficient of $x^4$ and $x$ ,but ended up failing miserably...

I began to think it is a function that has no real roots because it just makes me feel better to think of it that way! But seriously though, I think it's a doable problem, but it's just that my ability to solve it is limited.

Could anyone show me the correct approach to crack it?

Thanks in advance.

Problem:

Find all real roots of the quartic $x^4-(2k + 1)x^2 -x + k^2+ k - 1 = 0$ correct to 4 decimal places, where $k= 10^{10}$.
My two cents:

If we assume that all of the roots are much smaller than \(\displaystyle k^2\) then we can approximate the equation as
\(\displaystyle x^4 - (2k + 1)x^2 + k^2 = 0\)

This is a biquadratic equation; Use \(\displaystyle y = x^2\), (which gives \(\displaystyle y^2 - (2k + 1)y + k^2 = 0\)), solve for y using the quadratic formula, then take the square root to get it back in terms of x. It turns out that my assumption is a good one.

-Dan
 

paulmdrdo

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May 13, 2013
386
is this a pre-algebra or algebra level problem?
 

MarkFL

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Feb 24, 2012
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is this a pre-algebra or algebra level problem?
Yes, although this problem may be more difficult than that which would typically be encountered in an elementary algebra course, finding the roots of polynomials is a topic in algebra.
 
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anemone

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Feb 14, 2012
3,753
Let $f(x) = x^4-(2k + 1)x^2 -x + k^2+ k - 1$ (where for the moment $k$ is just a large positive number). Then $f(k^{1/2}) = -k^{1/2} - 1 < 0$. But $f$ is positive at $x=0$ and also as $x\to\infty$. So $f$ must have a real root between $0$ and $k^{1/2}$, and another real root greater than $k^{1/2}$.

My next move would be to let $x=k^{1/2}+y$ and get an equation for $y$. I would hope to get some information indicating that $y$ is approximately $\pm k^\alpha$, for some value of $\alpha$ less than $1/2$. But I don't have time to do that just now.
Thank you Opalg for the reply...

I see that if I let $x=k^{1/2}+y$, then I get

$f(x)=f(k^{1/2}+y)$

$\;\;\;\;\;\;\;=(k^{1/2}+y)^4-(2k+1)(k^{1/2}+y)^2-(k^{1/2}+y)+k^2+k-1$

$\;\;\;\;\;\;\;=y^4+4\sqrt{k}y^3+(4k-1)y^2-(2\sqrt{k}+1)y-\sqrt{k}$

And if I let $y=1$, I get

$f(x)=f(k^{1/2}+1)=1+4\sqrt{k}+4k-1-2\sqrt{k}-1-\sqrt{k}=4k+\sqrt{k}-1>0$

Hence, another root will be in between $\sqrt{k}$ and $\sqrt{k}+1$, i.e. 100,000 and 100,001.

That is the far that I can get, Opalg, without finding the real roots of them.

My two cents:

If we assume that all of the roots are much smaller than \(\displaystyle k^2\) then we can approximate the equation as
\(\displaystyle x^4 - (2k + 1)x^2 + k^2 = 0\)

This is a biquadratic equation; Use \(\displaystyle y = x^2\), (which gives \(\displaystyle y^2 - (2k + 1)y + k^2 = 0\)), solve for y using the quadratic formula, then take the square root to get it back in terms of x. It turns out that my assumption is a good one.

-Dan
Hey Dan,

Thank you for the reply...I understand of your approach and you have pointed out something very insightful to me, that is to leave out all other insignificant values when we're dealing with the addition and/or subtraction of some larger values. :)

is this a pre-algebra or algebra level problem?
This is an algebra problem, Paul!:eek:
 
Last edited:

Opalg

MHB Oldtimer
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Feb 7, 2012
2,725
I see that if I let $x=k^{1/2}+y$, then I get

$f(x)=f(k^{1/2}+y)$

$\;\;\;\;\;\;\;=(k^{1/2}+y)^4-(2k+1)(k^{1/2}+y)^2-(k^{1/2}+y)+k^2+k-1$

$\;\;\;\;\;\;\;=y^4+4\sqrt{k}y^3+(4k-1)y^2-(2\sqrt{k}+1)y-\sqrt{k}$
Since $k$ is going to be enormously larger than $y$, I prefer to write this in powers of $k$, namely $$f(k^{1/2}+y) = 4y^2k + (4y^3-2y-1)k^{1/2} + (y^4-y^2-y-1).$$
And if I let $y=1$, I get

$f(x)=f(k^{1/2}+1)=1+4\sqrt{k}+4k-1-2\sqrt{k}-1-\sqrt{k}=4k+\sqrt{k}-1>0$
You can take $y$ very much smaller than that. For example, if $y = -\frac12k^{-1/4}$ then you find that $f(k^{1/2}+y) = k^{1/4} -1 -\frac14k^{-1/2} + \frac1{16}k^{-1} > 0$, so (putting $k=10^{10}$) there is a root between $99\,999.9971883$ and $100\,000$. Also, the derivative $f'(x)$ is approximately $-4k^{3/4}$ at that point, which is large enough to ensure that the root occurs between $99\,999.99718$ and $99\,999.99719$. So the smaller root should be $99\,999.9972$ to four decimal places.

The larger root will also be extraordinarily close to $100\,000$, and I think it will be $100\,000.0028$ to four decimal places.
 

Prove It

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MHB Math Helper
Jan 26, 2012
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anemone

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Feb 14, 2012
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Since $k$ is going to be enormously larger than $y$, I prefer to write this in powers of $k$, namely $$f(k^{1/2}+y) = 4y^2k + (4y^3-2y-1)k^{1/2} + (y^4-y^2-y-1).$$
Ops...I think this is one of the unforgivable mistake that I made, given I have read what topsquark has suggested...(Angry)


You can take $y$ very much smaller than that. For example, if $y = -\frac12k^{-1/4}$ then you find that $f(k^{1/2}+y) = k^{1/4} -1 -\frac14k^{-1/2} + \frac1{16}k^{-1} > 0$, so (putting $k=10^{10}$) there is a root between $99\,999.9971883$ and $100\,000$. Also, the derivative $f'(x)$ is approximately $-4k^{3/4}$ at that point, which is large enough to ensure that the root occurs between $99\,999.99718$ and $99\,999.99719$. So the smaller root should be $99\,999.9972$ to four decimal places.

The larger root will also be extraordinarily close to $100\,000$, and I think it will be $100\,000.0028$ to four decimal places.
Hmm...I think there is a miscalculation to the $x$ value because I think $99\,999.99718$ should be $99\,999.99842$.

I see I see...I understand it now, Opalg!

Since we have $f(\sqrt{k}+0)<0$, we need to pick one negative $y$ value that is so small such that when we subtracted it from $\sqrt{k}$, it will give us

i) $f(\sqrt{k}+y)>0$,

ii) a value $(\sqrt{k}+y)$ that deviates only slightly from $10000$.

We could work further from that to find an educated guess for another negative $y$ value to deduce the smaller positive root for $x$, correct to 4 decimal places, as wanted.

And thank you for given me the first good estimation of the $y$ value, i.e. $y = -\frac12k^{-1/4}=-0.001581$...

And below is my working to find the smaller positive root of the equation:

$\small f(10^{5}-0.001581) = 4(0.001581)^2(10^{10}) + (4(0.001581)^3-2(0.001581)-1)10^{5} + ((0.001581)^4-(0.001581)^2-(0.001581)-1)>0$

$\small f(10^{5}-0.001579) = 4(0.001579)^2(10^{10}) + (4(0.001579)^3-2(0.001579)-1)10^{5} + ((0.001579)^4-(0.001579)^2-(0.001579)-1)>0$

$\small f(10^{5}-0.001578) = 4(0.001578)^2(10^{10}) + (4(0.001578)^3-2(0.001578)-1)10^{5} + ((0.001578)^4-(0.001578)^2-(0.001578)-1)<0$

Hence, we can say that there is a root in between $(99999.998419, 99999.998422)$ and we can conclude by now the smaller positive root, correct to 4 decimal places is $x=99999.9984$.:)

You can use this method:

View attachment 1488
Thank you so much for the pdf file, Prove It!:)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Hmm...I think there is a miscalculation to the $x$ value because I think $99\,999.99718$ should be $99\,999.99842$.
Ahem, yes. The trouble, as so often, is that I can't do basic arithmetic. I wanted to find $\bigl(10^{10}\bigr)^{1/4}$ on my calculator, and I made the mistake of thinking that $10\times\frac14 = 2.25.$ (Blush) (Fubar)

The answer should indeed be $99\,999.99842$, and the larger root should be $100\,000.00158$.
 

bergausstein

Active member
Jul 30, 2013
191
anemone, this kind of problem is interesting. would you be kind to tell me from what algebra book did you get this problem? because the one i'm currently using contain boring exercise.
 

anemone

MHB POTW Director
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Feb 14, 2012
3,753
anemone, this kind of problem is interesting. would you be kind to tell me from what algebra book did you get this problem? because the one i'm currently using contain boring exercise.
I'm very sorry to tell you that all of the problems that I post here, whether they are challenge problems or problems with which I need to ask for help come from a collection of problems that I have written out in my book. And to be completely honest with you, I have to then try to recall from memory where I copied them from if someone helping me asks for the source of the problem, with no guarantee of success.