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Real roots

Prove that the roots of the equation can't be all real if \(2a^2 <5b^2\) :

\[x^5+ax^4+bx^3+cx^2+dx+e\]

Note by Ankit Kumar Jain
7 months ago

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SOLUTION 2:

Lemma : For any equation \(f(x)\) of degree \(n\) to have all it's roots real , it must be true that all the roots of \(f^{r}(x)\) must also have it's all roots real , where \(1 \leq r \leq n\). BUT THE CONVERSE IS NOT TRUE.

Proof : If you see this graphically . we observe that if all the roots of \(f(x)\) of degree \(n\) are real , then it must be true that the graph of the polynomial must take turns \(n-1\) times or we can say that \(f^{'}(x)\) must be \(0\) for \(n-1\) values including multiplicity of roots.

So proceeding this way leads us to the desired result.


So, if the equation in the question has all it's roots real , then it's third derivative must also have all it's roots real i.e. \(60x^2 +24ax + 6b = 0\) has all it's roots real.

\(\Rightarrow D \geq 0 \Rightarrow \boxed{2a^2 \geq 5b}\).

But the conditions provided in the question are contrary to what is an essential for all the roots to be real.Hence , all the roots of the equation can't be real.

Ankit Kumar Jain - 7 months ago

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Thank you very much (+1)

Rahil Sehgal - 7 months ago

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:) :)

Ankit Kumar Jain - 7 months ago

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Let \( x_{1},x_{2} \cdots x_{5} \) be the real roots of the polynomial \( P(x) \)

Then \( x_{1} + x_{2} + \cdots + x_{5} = S = -a \)

Let \( \sum x_{1} \cdot x_{2} = T \)

Then you can directly prove by applying AM -GM twice

Rahil Sehgal - 7 months ago

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I have summarized your solution as solution 1 to the problem ..See that..

Ankit Kumar Jain - 7 months ago

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Thanks... I was thinking to post it but a hint is actually enough.

Rahil Sehgal - 7 months ago

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@Rahil Sehgal If you did it something differently then you can post your solution too. And as for the solution 2 ..I am just posting it..wait..

Ankit Kumar Jain - 7 months ago

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Exactly..You got it correct..By the way..There is an algebraic proof also..I mean a proof using no inequalities.

Ankit Kumar Jain - 7 months ago

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I will try to prove it without inequalities....

Btw nice question. Have u made it yourself?

Rahil Sehgal - 7 months ago

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@Rahil Sehgal No.

Ankit Kumar Jain - 7 months ago

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@Tapas Mazumdar@Kushal Bose@Akshat Sharda Post your solutions guys.

Ankit Kumar Jain - 6 months, 4 weeks ago

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@Md Zuhair @Aditya Narayan Sharma @Anirudh Sreekumar@Brian Charlesworth@Pi Han Goh Please post your solutions!

It will help everyone know alternate solutions to the problem,

Ankit Kumar Jain - 6 months, 4 weeks ago

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SOLUTION 1:

Suppose that the roots of the equation are \(x_1,x_2,x_3,x_4,x_5\) , all \(\in \mathbb{R}\)

By AM-GM Inequality , we have \(x_i^2+x_j^2 \geq 2x_ix_j\) .

Writing similarly and adding for all pairs of \((i,j)\) such that \(1 \leq i < j \leq 5\).

We get \(4\displaystyle \sum_{i=1}^{5} x_{i}^2 \geq 2\displaystyle \sum_{1 \leq i < j \leq 5} x_ix_j\)

Adding \(8\displaystyle \sum_{1 \leq i < j \leq 5} x_ix_j\) both sides we get

\(4\left(\displaystyle \sum_{i=1}^{5} x_i\right)^2 \geq 10\displaystyle \sum_{1 \leq i < j \leq 5} x_ix_j\)

Using Vieta's Relations : \(\displaystyle \sum_{i=1}^{5} x_i = a , \displaystyle \sum_{1 \leq i < j \leq 5} x_ix_j = b\)

Therefore , we get \(\boxed{2a^2 \geq 5b}\).

But the condition given in the question is just the contrary to what is an essential for all roots to be real..Hence the equation can't have all it's roots real under the given condition of \(2a^2 < 5b\)

Ankit Kumar Jain - 7 months ago

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Are you assuming that \(x_i\) are all real roots here? If that's not the case, then you cannot apply AM-GM at all as \({x_i}^2\) can be negative. I'm assuming this is a proof by contradiction but for that first you need to highlight the point \(x_i\) are all real.

Tapas Mazumdar - 6 months, 2 weeks ago

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Thanks!...I have edited the solution. :)

Ankit Kumar Jain - 6 months, 1 week ago

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@Rahil Sehgal I have fixed a minor issue here...The symbol should be greater than equal to in that box..I had earlier mentioned a strict inequality there.

Ankit Kumar Jain - 7 months ago

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Thanks... I didn't notice that much.

Rahil Sehgal - 7 months ago

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@Rahil Sehgal You can try this ...Inequality is back..I have posted that today itself. And even this Algebraic Manipulation ..you can see that discussion link in the contributions tab in my profile.

Ankit Kumar Jain - 7 months ago

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@Ankit Kumar Jain OK. Sure.

Rahil Sehgal - 7 months ago

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Can you please post the solution 2 ( without inequalities).

This method is actually the elaboration of my solution.

Rahil Sehgal - 7 months ago

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Oh..yes sorry..I used the wrong word 'summarized' , it should be 'elaborated'.

Ankit Kumar Jain - 7 months ago

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@Rahil Sehgal Here is an inequality problem..

Ankit Kumar Jain - 7 months ago

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Do we need to find the discriminant (D) of the equation to prove this?

If D ≥ 0 then only roots are real.

Rahil Sehgal - 7 months ago

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You can use this ...Consider that all the roots of the equation are real..Then try to introduce some inequalities to get the desired result.

Ankit Kumar Jain - 7 months ago

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But there is no discriminant for a degree 5 equation ..as far as I know..Tell me if I am wrong.

Ankit Kumar Jain - 7 months ago

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@Ankit Kumar Jain We can actually find it... See the wiki. I found this.

For a polynomial \(P(x)=a_n x^n+a_{n-1} x^{n-1}+\cdots+a_1 x+a_0\) having roots \(x_1,x_2,\ldots,x_n\) (counting multiplicity), its discriminant is: \[\Delta=a_n^{2n-2}\prod_{1 \leq i < j \leq n} (x_i-x_j)^2\]

Rahil Sehgal - 7 months ago

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@Rahil Sehgal What is the title of the wiki?

Ankit Kumar Jain - 7 months ago

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@Ankit Kumar Jain See this

Rahil Sehgal - 7 months ago

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@Rahil Sehgal Thanks!

Ankit Kumar Jain - 7 months ago

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@Rahil Sehgal See this

Rahil Sehgal - 7 months ago

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@Rahil Sehgal That is good...But will that help here?

Ankit Kumar Jain - 7 months ago

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