Question :

- \( x^4 + px^3 + qx^2 + px + 1 =0 \) 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 ?

## Comments

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TopNewestHey, 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 · 4 months, 2 weeks ago

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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 · 4 months, 2 weeks ago

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– Yatin Khanna · 4 months, 2 weeks ago

Thanks :)Log in to reply

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? – Calvin Lin Staff · 4 months, 2 weeks ago

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– Anurag Pandey · 4 months, 2 weeks 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 ?Log in to reply

Think through all the possible cases and their implications. – Calvin Lin Staff · 4 months, 2 weeks ago

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– Anurag Pandey · 4 months, 2 weeks 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 .Log in to reply

@Calvin Lin Thank you Sir ! – Anurag Pandey · 4 months, 2 weeks ago

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– Calvin Lin Staff · 4 months, 2 weeks ago

Can you add your complete solution? Reading your comments, I'm not certain that you have the correct condition to check.Log in to reply