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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
4 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. · 4 months ago

Thank you very much (+1) · 4 months ago

:) :) · 4 months ago

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 · 4 months ago

I have summarized your solution as solution 1 to the problem ..See that.. · 4 months ago

Thanks... I was thinking to post it but a hint is actually enough. · 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.. · 4 months ago

Exactly..You got it correct..By the way..There is an algebraic proof also..I mean a proof using no inequalities. · 4 months ago

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

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

No. · 4 months ago

@Tapas Mazumdar@Kushal Bose@Akshat Sharda Post your solutions guys. · 3 months, 4 weeks ago

It will help everyone know alternate solutions to the problem, · 3 months, 4 weeks ago

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 $$2a^2 < 5b$$ · 4 months ago

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. · 3 months, 2 weeks ago

Thanks!...I have edited the solution. :) · 3 months, 1 week ago

@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. · 4 months ago

Thanks... I didn't notice that much. · 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. · 4 months ago

OK. Sure. · 4 months ago

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

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

Oh..yes sorry..I used the wrong word 'summarized' , it should be 'elaborated'. · 4 months ago

@Rahil Sehgal Here is an inequality problem.. · 4 months ago

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

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

But there is no discriminant for a degree 5 equation ..as far as I know..Tell me if I am wrong. · 4 months ago

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: $\Delta=a_n^{2n-2}\prod_{1 \leq i < j \leq n} (x_i-x_j)^2$ · 4 months ago

What is the title of the wiki? · 4 months ago

See this · 4 months ago

Thanks! · 4 months ago

See this · 4 months ago