Polynomial Doubt

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 ?

#Algebra

Easy Math Editor

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Comments

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 years, 5 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 ?

Anurag Pandey - 4 years, 5 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.

Calvin Lin Staff - 4 years, 5 months ago

@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 $x=\pm{2}$ will be greater than or equal to zero .

Anurag Pandey - 4 years, 5 months ago

@Anurag Pandey – @Calvin Lin Thank you Sir !

Anurag Pandey - 4 years, 5 months ago

@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 - 4 years, 5 months ago

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 - 4 years, 5 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.

Anurag Pandey - 4 years, 5 months ago

Thanks :)

Yatin Khanna - 4 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
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`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$