Polynomial Doubt

Question :

  1. x4+px3+qx2+px+1=0 x^4 + px^3 + qx^2 + px + 1 =0 has real roots. Then what is the minimum value of p2+q2 p^2 +q^2 .

How I started ?
I started by dividing the whole equation by x2x^2 then we get
(x+1x)2+p(x+1x)+q2=0 (x + \frac{1}{x} ) ^2 + p (x + \frac{1}{x} ) + q - 2 = 0
Then put (x+1x)=t(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
2 years, 5 months ago

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1 vote

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Great start! So what you have is: The quadratic equation t2+pt+q2=0 t^2 + pt + q-2 = 0 must have 2 roots that are not in (2,2) (-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?

Calvin Lin Staff - 2 years, 5 months ago

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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 ?

Anurag Pandey - 2 years, 5 months ago

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No necessarily. (x100)2 ( x - 100)^2 satisfies the requirements, but the value at 2, -2 is positive.

Think through all the possible cases and their implications.

Calvin Lin Staff - 2 years, 5 months ago

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

Anurag Pandey - 2 years, 5 months ago

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@Anurag Pandey @Calvin Lin Thank you Sir !

Anurag Pandey - 2 years, 5 months ago

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@Anurag Pandey Can you add your complete solution? Reading your comments, I'm not certain that you have the correct condition to check.

Calvin Lin Staff - 2 years, 5 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?

Yatin Khanna - 2 years, 5 months ago

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p2>4(q2)p^2 > 4(q-2) has to be true for the quadratic to have real roots. But now as we took t=x+1x 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.

Anurag Pandey - 2 years, 5 months ago

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Thanks :)

Yatin Khanna - 2 years, 5 months ago

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