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

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

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

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

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TopNewestSOLUTION 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 · 4 months ago

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– Rahil Sehgal · 4 months ago

Thank you very much (+1)Log in to reply

– Ankit Kumar Jain · 4 months ago

:) :)Log in to reply

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 · 4 months ago

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– Ankit Kumar Jain · 4 months ago

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

– Rahil Sehgal · 4 months ago

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

– Ankit Kumar Jain · 4 months ago

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..Log in to reply

– Ankit Kumar Jain · 4 months ago

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

Btw nice question. Have u made it yourself? – Rahil Sehgal · 4 months ago

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– Ankit Kumar Jain · 4 months ago

No.Log in to reply

@Tapas Mazumdar@Kushal Bose@Akshat Sharda Post your solutions guys. – Ankit Kumar Jain · 3 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 · 3 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 · 4 months ago

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\(x_i\) are all real. – Tapas Mazumdar · 3 months, 2 weeks agoLog in to reply

– Ankit Kumar Jain · 3 months, 1 week ago

Thanks!...I have edited the solution. :)Log in to reply

@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 · 4 months ago

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– Rahil Sehgal · 4 months ago

Thanks... I didn't notice that much.Log in to reply

– Ankit Kumar Jain · 4 months ago

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.Log in to reply

– Rahil Sehgal · 4 months ago

OK. Sure.Log in to reply

This method is actually the elaboration of my solution. – Rahil Sehgal · 4 months ago

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– Ankit Kumar Jain · 4 months ago

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

@Rahil Sehgal Here is an inequality problem.. – Ankit Kumar Jain · 4 months ago

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If D ≥ 0 then only roots are real. – Rahil Sehgal · 4 months ago

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– Ankit Kumar Jain · 4 months ago

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.Log in to reply

– Ankit Kumar Jain · 4 months ago

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

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 · 4 months ago

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– Ankit Kumar Jain · 4 months ago

What is the title of the wiki?Log in to reply

this – Rahil Sehgal · 4 months ago

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– Ankit Kumar Jain · 4 months ago

Thanks!Log in to reply

this – Rahil Sehgal · 4 months ago

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– Ankit Kumar Jain · 4 months ago

That is good...But will that help here?Log in to reply