A Problem solved wrong By Pi han goh and Calvin Lin

Let $P(x)=x^{3}-3b^{2}x+16$. If $P(|x|)$ has exactly two real roots, then the area bounded by $y=P(x)$ and $x$-axis is?

The answer is infinite. Because if $ω$ is a root of $P(|x|)$ then $-\omega$ is also a root of $P(|x|)$. This means $P(x)$ can have only one real root either $\omega$ or $-\omega$.This means that $P(x)$ intersects the $x$-axis once. So area bounded by it and $x-axis$ must be infinite.

This question was posted by me . And Pi Han goh and Calvin lin reported it and deleted it saying answer is not always infinity. But answer is always infinite .

Please tell me if I am right . Note by Shivam Jadhav
3 years, 4 months ago

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I disagree with " $P(|x|)$ has exactly two real roots means that $P(x)$ can have only one real root". What is true is

This mean $P(x)$ has exactly one positive root (and two negative roots if we want $P(x)$ to have 3 real roots.).

As an explicit example, we could have $P( -10 ) = P(-4) = P(6) = 0$. Then if we solve for $P(|x|) = 0$, we only get the roots $x = 6, -6$.

Do you see the error in your argument? When you said "because if $\omega$ is a root", what should that actually be?

Please be mindful of your title. I generally have a very good reason if I believe that something is wrong. It would be better for you to ask "Why was this problem deleted" instead of claiming that "Calvin Lin solved this wrongly". Of course, I do admit that there are times that I solve the problem / resolve the report incorrectly, because I'm speed reading through it / missed a condition.

Staff - 3 years, 4 months ago

Sir but here product of roots is $-16$ . So the case stated by you doesn't satisfy.

- 3 years, 4 months ago

Yes I know. I was only responding to the specific condition of "If $P(|x|)$ has exactly two real roots, what can we conclude?", and your conclusion of "$P(x)$ can have only one real root" is false.

Note that there is no explicit counter example to the entire problem, because there are no polynomials which satisfy the conditions that you stated, namely
1. $P(x) = x^3 - 3b^2 x + 16$
2. $P(|x|)$ has 2 real roots counted with multiplicity.

Reasoning: From condition 1, since $P(0) > 0$, and we have a leading coefficient of 1, the polynomial either has 2 positive real roots (with multiplicity) or 0 positive real roots. As explained above, condition 2 implies that there is exactly 1 positive real root.

Thus, that's another reason for deletion, since your problem is about the empty set.

Staff - 3 years, 4 months ago

That's right but considering the question as logical not the calculations one the question can be valid .

- 3 years, 4 months ago

How is the question valid? For what value of $b$ are your conditions satisfied?

Staff - 3 years, 4 months ago

Substitute $b = \pm 2$, the answer is 24.

- 3 years, 4 months ago

When $b=2,-2$ $P(|x|)$ has three real roots $2,2,-2$ and as $2$ is a repeated root we consider that $P(|x|)$ has three real roots with $2$ as the repeated root. And in question it is strictly mentioned that $P(|x|)$ has exactly two real roots. So please notice this and tell that whether I am right or wrong .

- 3 years, 4 months ago

Then you should mention that we're counting the multiplicity of the roots.

And are you counting complex roots as well? If so, you need to make it explicit.

- 3 years, 4 months ago

We are taking only real roots . It's mentioned in the question

- 3 years, 4 months ago

Then why is $\omega$, a third root of unity in your solution?

- 3 years, 4 months ago

$ω$ is a variable

- 3 years, 4 months ago

So what is the value of $b$ that gives the answer of infinity?

- 3 years, 4 months ago

@Pi Han Goh @Calvin Lin

- 3 years, 4 months ago

My understanding of cubics is that if it has a complex root, ia + ib, then a - ib is also a root, so a cubic always has 1 real root or 3 real roots. so the question is meaningless. Ed Gray

- 1 year, 5 months ago

@Shivam Jadhav, knowing Pi Han Goh and Calvin Lin, I don't think they could possibly get anything incorrect.

- 1 year, 11 months ago