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# Polynomial Doubt

Question :

1. $$x^4 + px^3 + qx^2 + px + 1 =0$$ has real roots. Then what is the minimum value of $$p^2 +q^2$$ .

How I started ?
I started by dividing the whole equation by $$x^2$$ then we get
$$(x + \frac{1}{x} ) ^2 + p (x + \frac{1}{x} ) + q - 2 = 0$$
Then put $$(x + \frac{1}{x} ) = t$$. Then discriminant should be greater than equal to zero. But now the problem arises that t does not belong to (-2,2) , so taking care of that part leads to solving inequality which I am unable to do .

Have I started the right way?
One more thing to notice is that the sum of roots of the equation is equal to the sum of reciprocal of the roots .
How to proceed further ?

Note by Anurag Pandey
1 year, 2 months ago

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Hey, can you help me in solving the case when p^2>4(q-2)? How do we obtain the minimum value for this case?

- 1 year, 2 months ago

$$p^2 > 4(q-2)$$ has to be true for the quadratic to have real roots. But now as we took $$t = x + \frac{1}{x}$$ whose value does not lie in (-2,2) so we have to make sure that |t| is greater than or equal to |2|.
I used the condition stated below in comment. And then we get an inequality. My friend further used Cauchy Schwarz inequality to find the minimum value.
Hope this helps.

- 1 year, 2 months ago

Thanks :)

- 1 year, 2 months ago

Great start! So what you have is: The quadratic equation $$t^2 + pt + q-2 = 0$$ must have 2 roots that are not in $$(-2,2)$$. What are the necessary and sufficient condtions for this to happen?

For example, you stated that "discriminant must be positive/non-negative", which is clearly necessary, but not sufficient. How can we check to ensure that the roots are in the desired regions?

Staff - 1 year, 2 months ago

As the coefficient of ' t ' is positive so the value of the quadratic at x=-2 ans at x =2 must be less than or equal to zero. Equality can hold as it can take value -2 and 2. Is this condition sufficient ?

- 1 year, 2 months ago

No necessarily. $$( x - 100)^2$$ satisfies the requirements, but the value at 2, -2 is positive.

Think through all the possible cases and their implications.

Staff - 1 year, 2 months ago

Okay . I forgot to take in account the case where discriminant is equal to zero. In that case the value of quadratic at $$x=\pm{2}$$ will be greater than or equal to zero .

- 1 year, 2 months ago

@Calvin Lin Thank you Sir !

- 1 year, 2 months ago

Staff - 1 year, 2 months ago

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