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.

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 \(2a^2 < 5b\)

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.

@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.

@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
–
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:
\[\Delta=a_n^{2n-2}\prod_{1 \leq i < j \leq n} (x_i-x_j)^2\]

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## Comments

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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.

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Thank you very much (+1)

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:) :)

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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

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

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

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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.

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

Btw nice question. Have u made it yourself?

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

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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,

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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 \(2a^2 < 5b\)

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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.Log in to reply

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

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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.

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

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

This method is actually the elaboration of my solution.

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

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@Rahil Sehgal Here is an inequality problem..

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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.

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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.

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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.

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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: \[\Delta=a_n^{2n-2}\prod_{1 \leq i < j \leq n} (x_i-x_j)^2\]

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this

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this

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