MHB Solve x^3-3x=\sqrt{x+2}: Get Help Here!

[anemone]

Hi all,

I've tried to no avail to find the solution to this equation in exact form, and I'm giving it up entirely and I hope someone would come up with a brilliant idea on how to tackle it elegantly.

Thanks in advance.:)

Problem: Solve $$x^3-3x=\sqrt{x+2}$$.

 

[topsquark]

Hi all,

I've tried to no avail to find the solution to this equation in exact form, and I'm giving it up entirely and I hope someone would come up with a brilliant idea on how to tackle it elegantly.

Thanks in advance.:)

Problem: Solve $$x^3-3x=\sqrt{x+2}$$.

Please show some work!

Hint: By the rational root theorem you have possible rational solutions: \pm 1, \pm 2. Do any of these work?

-Dan

 

[Sudharaka]

Hi all,

I've tried to no avail to find the solution to this equation in exact form, and I'm giving it up entirely and I hope someone would come up with a brilliant idea on how to tackle it elegantly.

Thanks in advance.:)

Problem: Solve $$x^3-3x=\sqrt{x+2}$$.

Hi anemone, :)

Just to add something to what Dan has suggested, after finding a rational solution you can use polynomial long division to find the other solutions of this equation. :)

 

[Petrus]

Please show some work!

Hint: By the rational root theorem you have possible rational solutions: \pm 1, \pm 2. Do any of these work?

-Dan

Hello,
I am just curious how this rational root theorem works when we got a square root in the constant
We got
$$-x^3+3x+\sqrt{x+2}=0$$
I don't understand how I will get my constant term? If we look at this I know that our $$a_n=-1$$ but what is our $$a_0$$

Regards,
$$|\pi\rangle$$

 

[Prove It]

Hi all,

I've tried to no avail to find the solution to this equation in exact form, and I'm giving it up entirely and I hope someone would come up with a brilliant idea on how to tackle it elegantly.

Thanks in advance.:)

Problem: Solve $$x^3-3x=\sqrt{x+2}$$.

Did you think to maybe square both sides to remove the square root and give you a polynomial which you actually have a chance of being able to solve?

 

[alyafey22]

Hello,
I am just curious how this rational root theorem works when we got a square root in the constant
We got
$$-x^3+3x+\sqrt{x+2}=0$$
I don't understand how I will get my constant term? If we look at this I know that our $$a_n=-1$$ but what is our $$a_0$$

Regards,
$$|\pi\rangle$$

You have got to first square both sides of the equation .

 

[kaliprasad]

Hi all,

I've tried to no avail to find the solution to this equation in exact form, and I'm giving it up entirely and I hope someone would come up with a brilliant idea on how to tackle it elegantly.

Thanks in advance.:)

Problem: Solve $$x^3-3x=\sqrt{x+2}$$.

If we square both sides( I am not squaring to find the expression) then
we get x 6th order polynomial the coefficient of x^6 is 1 and constant term of is -2 as x+2 goes to left
so if it has a rational root it is factor of -2 that is 1/-1/2/-2
1 and -1 can be ruled out as they give give irrational so checking - 2 and 2 we get that x= 2 satisfies both sides
hence it is a root.

to find all solutions is a different matter ( others are irrational)

 

[SuperSonic4]

Hi all,

I've tried to no avail to find the solution to this equation in exact form, and I'm giving it up entirely and I hope someone would come up with a brilliant idea on how to tackle it elegantly.

Thanks in advance.:)

Problem: Solve $$x^3-3x=\sqrt{x+2}$$.

I would first note that because of the square root the domain of this equation is $x+2 \geq 0 \ \therefore \ x \geq -2$

I would then square both sides to eliminate the square root
$(x^3-3x)^2 =(\sqrt{x+2})^2$

Then use the rational roots theorem on the 6th degree polynomial to check for an integer solution (and hence a factor)

Spoiler

$x^6 - 6x^4 + 9x^2 - x - 2 = 0$

This is not a nice equation to solve

hint: f(2) = 0

 

[Petrus]

Hello,
Thanks for the fast responed! It's make sense now

Regards,
$$|\pi\rangle$$

 

[Farmtalk]

Hi anemone, :)

Just to add something to what Dan has suggested, after finding a rational solution you can use polynomial long division to find the other solutions of this equation. :)

That would be through the use of Descartes' rule of signs, correct?

 

[topsquark]

That would be through the use of Descartes' rule of signs, correct?

No. You know that x = 2 solves the equation. Thus the original polynomial (the x^6 + ...) contains a factor of x - 2. So divide the polynomial by x - 2. (Or use synthetic division. Whatever.) That will give you the next polynomial to factor.

Work it through and let us know what you get.

-Dan

 

[Sudharaka]

That would be through the use of Descartes' rule of signs, correct?

Hi Farmtalk, :)

Here's how I thought about it (Without Descartes' rule of signs)...

Spoiler

\[x^3-3x=\sqrt{x+2}\]

\[\Rightarrow x^6-6x^4+9x^2-x-2=0\]

From the rational root theorem we get \(x=2\) as a solution. So by polynomial long division we get,

\[(x-2)(x^5+2x^4-2x^3-4x^2+x+1)=0\]

If \(x^5+2x^4-2x^3-4x^2+x+1\) is reducible it should be the product of a quadratic and a cubic. Otherwise there should be another rational root. Using a little trial and error (trying to factor the above quintic with monic polynomials) I got,

\[x^5+2x^4-2x^3-4x^2+x+1=(x^2+x-1)(x^3+x^2-2x-1)\]

Hence,

\[x^6-6x^4+9x^2-x-2=(x-2)(x^2+x-1)(x^3+x^2-2x-1)\]

Note that we have to discard the solutions of the above, that doesn't satisfy our original equation.

 

[Farmtalk]

If I could give you 2 thanks I would! Much appreciated! :D

I think I've learned more math in the last couple weeks on here than in a couple maths of my classes

 

[anemone]

Thanks everyone for the replies...obviously I didn't express myself clearly enough in this thread because:

1) I should have said "Find the solutions to this equation" instead of just a solution as I stated in my first post. Yes, I've found and verified that $$x=2$$ is one of the roots of the given equation.

2) I really should include all that I've tried and where I struggled so that I would save the time of the helpers in their quest to point out to me my potential errors and to give suggestions/hints to me on how to possibly proceed.

3) I should tell that I've plotted the two graphs $$y=x^3-3x$$ and $$y=\sqrt{x+2}$$ on the same Cartesian plane and I knew there are two negative real roots and one positive real root to this problem.

View attachment 943

Also, I'm sorry for the delay in responding...

Hi Sudharaka,

Thank you so much for factoring this out for me; I deeply appreciate that!

To find the real roots from:

$$x^6-6x^4+9x^2-x-2=(x-2)(x^2+x-1)(x^3+x^2-2x-1)$$

we see that $$x=2$$ is one of the real roots.

And for $$x^2+x-1=0$$, we find

$$x=\frac{-1\pm \sqrt{5}}{2}=\frac{-1-\sqrt{5}}{2}\;\text{or} \frac{-1\pm \sqrt{5}}{2}$$

Checking these roots to see which, if any, are extraneous, we find:

$$x=-\frac{1+ \sqrt{5}}{2}$$ is another real root.

For the third factor, I approached it by using the Newton-Raphson method, and obtained:

$$f(x)=x^3+x^2-2x-1,\;\;f(-0.45)=0.011372,\;\;f(-0.44)=-0.011584$$ and $$f'(x)=3x^2+2x-2$$.

Let the initial estimate of the root be $$x_0=-0.445$$

Thus,

$$x_1=x_0-\frac{f(x_0)}{f'(x_0)}=-0.445-\frac{-0.000096125}{-2.295925}=-0.445041867$$

$$x_2=x_1-\frac{f(x_1)}{f'(x_1)}=-0.445041867-\frac{-2.095301692205461363\times 10^{-9}}{-2.295896943851462933}=-0.445041867$$

Hence, the third real root is approximated by $$-0.445041867$$.

We can conclude now the real roots to this problem are

$$x=2,\;-\left(\frac{1+ \sqrt{5}}{2}\right)$$ and $$x\approx -0.445041867$$.

 

[anemone]

Hi MHB, I found the solution to this particular problem recently and I think it is only fair if I share it here, especially to Sudharaka who had helped me by posting so many valuable tips to me in this thread!

Problem: Solve $$x^3-3x=\sqrt{x+2}$$.

Solution. Helder Oliveira de CASTRO (ITA-Aeronautic Institute of Technology, Sao Paulo, Brazil) and Yufei ZHAO (Don Mills Collegiate Institute, Toronto, Canada, Grade 10).

If $x<-2$, then the right side of the equation is not defined. If $x>2$, then

$$x^3-3x=\frac{x^3+3x(x+2)(x-2)}{4}>\frac{x^3}{4}>\sqrt{x+2}$$

So the solution(s), if any, must be in $[-2,2]$. Write $x=2\cos(a)$, where $0\le a\le\pi$. The equation becomes

$$8\cos^3(a)-6\cos(a)=\sqrt{2\cos(a)+2}$$

Using the triple angle identity on the left side and the half angle identity on the right side, we get

$$2\cos(3a)=2\cos\left(\frac{a}{2} \right)\ge0$$

Then $$3a\pm\frac{a}{2}=2n\pi$$ for some integer $n$. Since $$3a\pm\frac{a}{2}\in\left[-\frac{\pi}{2},\frac{7\pi}{2} \right]$$, we get $n=0\text{ or }1$. We easily checked that $$a=0,\,\frac{4\pi}{5},\,\frac{4\pi}{7}$$ yield the only solutions $$x=2,\,2\cos\left(\frac{4\pi}{5} \right),\,2\cos\left(\frac{4\pi}{7} \right)$$.

 

[Sudharaka]

Hi MHB, I found the solution to this particular problem recently and I think it is only fair if I share it here, especially to Sudharaka who had helped me by posting so many valuable tips to me in this thread!

Problem: Solve $$x^3-3x=\sqrt{x+2}$$.

Solution. Helder Oliveira de CASTRO (ITA-Aeronautic Institute of Technology, Sao Paulo, Brazil) and Yufei ZHAO (Don Mills Collegiate Institute, Toronto, Canada, Grade 10).

If $x<-2$, then the right side of the equation is not defined. If $x>2$, then

$$x^3-3x=\frac{x^3+3x(x+2)(x-2)}{4}>\frac{x^3}{4}>\sqrt{x+2}$$

So the solution(s), if any, must be in $[-2,2]$. Write $x=2\cos(a)$, where $0\le a\le\pi$. The equation becomes

$$8\cos^3(a)-6\cos(a)=\sqrt{2\cos(a)+2}$$

Using the triple angle identity on the left side and the half angle identity on the right side, we get

$$2\cos(3a)=2\cos\left(\frac{a}{2} \right)\ge0$$

Then $$3a\pm\frac{a}{2}=2n\pi$$ for some integer $n$. Since $$3a\pm\frac{a}{2}\in\left[-\frac{\pi}{2},\frac{7\pi}{2} \right]$$, we get $n=0\text{ or }1$. We easily checked that $$a=0,\,\frac{4\pi}{5},\,\frac{4\pi}{7}$$ yield the only solutions $$x=2,\,2\cos\left(\frac{4\pi}{5} \right),\,2\cos\left(\frac{4\pi}{7} \right)$$.

Hi anemone, :)

Thanks for the solution. It sure is quite elegant. :)