What is going on here?

[Steven G]

This problem leads to a 4th degree polynomial with 4 real roots. Three of those roots were rejected as they were less than 3 + sqrt(3). The question is how did Newtons get a 2nd root using his method which was not any of the 4 roots from the 4th degree polynomial?

Thanks for looking!

 

[The Highlander]

Thanks for looking!

Is this you, Steven?

 

[Steven G]

Is this you, Steven?

No, that's Jomo!

 

[Otis]

This problem leads to a 4th degree polynomial with 4 real roots. Three of those roots were rejected as they were less than 3+sqrt(3)…

Hi Steven. Is that shown in an earlier video?

how did [Prime Newton] get a 2nd root which was not any of the 4 roots from the 4th degree polynomial?

That root could differ because he didn't solve a 4th-degree polynomial.

Also, Prime Newton arrives at his quadratic polynomial by a different line of reasoning than how you arrived at your quadratic polynomial.

Different approaches can lead to different extraneous solutions.

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In the video, you ask, "Why did I get a second answer?" I think it's because your approach had squared both sides of an equation, which often introduces extraneous solutions.

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You wonder also where a mistake might be in either your method or Prime Newton's method. I don't see any mistakes, but you didn't check your second solution. (It doesn't work because it's extraneous.)

Prime Newton eliminated one of his quadratic's roots at an earlier stage than you did, going on to arrive at the single solution shared by both of you.

 

[Otis]

I played with: [imath]\;\;x=n+\sqrt{n+\sqrt{x}}\quad n\in\mathbb{Z}^{+}[/imath]

The solution is: [imath]\;\;x=n+\frac{1}{2}+\frac{1}{2}\sqrt{4n+1}[/imath]

Does the expression below expanded give the 4th-degree polynomial that you'd referenced, when n=3?

\(\displaystyle \bigg(x-(n+\sqrt{n})\bigg)^2\bigg(x-(n-\sqrt{n})\bigg)^2-x\)

 

[Steven G]

Hi Steven. Is that shown in an earlier video?


That root could differ because he didn't solve a 4th-degree polynomial.

Also, Prime Newton arrives at his quadratic polynomial by a different line of reasoning than how you arrived at your quadratic polynomial.

Different approaches can lead to different extraneous solutions.

---

In the video, you ask, "Why did I get a second answer?" I think it's because your approach had squared both sides of an equation, which often introduces extraneous solutions.

---

You wonder also where a mistake might be in either your method or Prime Newton's method. I don't see any mistakes, but you didn't check your second solution. (It doesn't work because it's extraneous.)

Prime Newton eliminated one of his quadratic's roots at an earlier stage than you did, going on to arrive at the single solution shared by both of you.

It is very strange to me that doing a problem two different ways can lead to extraneous solutions. I have never come across that before. Are there any simple examples which you can share with me.

Yes, I never showed where the 4th degree polynomial came from.

 

[Steven G]

I played with: [imath]\;\;x=n+\sqrt{n+\sqrt{x}}\quad n\in\mathbb{Z}^{+}[/imath]

The solution is: [imath]\;\;x=n+\frac{1}{2}+\frac{1}{2}\sqrt{4n+1}[/imath]

Does the expression below expanded give the 4th-degree polynomial that you'd referenced, when n=3?

\(\displaystyle \bigg(x-(n+\sqrt{n})\bigg)^2\bigg(x-(n-\sqrt{n})\bigg)^2-x\)

Yes.

 

[Otis]

very strange to me that doing a problem two different ways can lead to extraneous solutions.

Ah, but that's not what I'd said.

If your extraneous root differs from my extraneous root, then that difference could be a result of you having used a different approach than I did. In other words, different approaches can lead to different extraneous solutions.

Prime Newton's quadratic polynomial differs from yours because his reasoning differs from yours. He discarded one of the roots. As he did not solve the referenced, quartic polynomial, I'm not surprised that the root he discarded is not one of the three discarded roots from solving the quartic polynomial. He reasoned differently, in obtaining his polynomial.

I hope that I haven't misinterpreted any of your questions.

 

[mario99]

It is very strange to me that doing a problem two different ways can lead to extraneous solutions. I have never come across that before. Are there any simple examples which you can share with me.

Yes, I never showed where the 4th degree polynomial came from.

Hi Professor Steven,

read the thread below, it has a simple example that leads to extraneous solutions.

Solving Radical Equations sqrt{x^2 - 2x + 1} = 2x + 5

(Solve the following radical equations. Be sure to identify any extraneous solutions.) sqrt{x^2 - 2x + 1} = 2x + 5 I’m supposed to get -4/3 but got -6. The way I did it is the inside part of the square root can be factored to (x-1)^2 so I just got rid of the square root without having to square...

www.freemathhelp.com

 

[Steven G]

Hi Professor Steven,

read the thread below, it has a simple example that leads to extraneous solutions.

Solving Radical Equations sqrt{x^2 - 2x + 1} = 2x + 5

(Solve the following radical equations. Be sure to identify any extraneous solutions.) sqrt{x^2 - 2x + 1} = 2x + 5 I’m supposed to get -4/3 but got -6. The way I did it is the inside part of the square root can be factored to (x-1)^2 so I just got rid of the square root without having to square...

www.freemathhelp.com

@mario99,
Yes, I know that an equation can have extraneous solutions. What I didn't know is that depending on how you solve an equation it is possible that you get different extraneous solutions.