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anemone
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let $a>b>c$ be roots of $\sqrt{2015}x^3-4031x^2+2=0$, find $ b(a+c)$.
anemone said:let $a>b>c$ be roots of $\sqrt{2015}x^3-4031x^2+2=0$, find $ b(a+c)$.
anemone said:Thanks kaliprasad for you good solution!:)
My solution:
Descartes' Rule of Signs tells us that this cubic polynomial has two positive roots and one negative root, hence, $c$ must be the negative root. . In fact, it has exactly three positive roots:
kaliprasad said:why the incorrect statement In fact, it has exactly three positive roots:
sorry for misunderstandinganemone said:Hi kaliprasad, isn't $\sqrt{2015}-\sqrt{2017}<0$ ?(Wink)
greg1313 said:\(\displaystyle f(x)=\sqrt{2015}x^3-4031x^2+2=0\implies x^3-\frac{4031}{\sqrt{2015}}x^2+\frac{2}{\sqrt{2015}}=0\)
By inspection, one of the roots is
\(\displaystyle \frac{1}{\sqrt{2015}}\)
(I must admit here that I looked at the other solutions before I came up with this).
\(\displaystyle f'(x)=3\sqrt{2015}x^2-8062x\)
The above is negative at \(\displaystyle x=\frac{1}{\sqrt{2015}}\), so we may conclude that all three roots are real and that \(\displaystyle b=\frac{1}{\sqrt{2015}}\).
From Vieta's formulae,
\(\displaystyle abc=\frac{-2}{\sqrt{2015}}\implies ac=-2\)
and
\(\displaystyle ab+ac+bc=0\implies b(a+c)=-ac=2\)
as required.
kaliprasad said:sorry for misunderstanding
In fact, it has exactly three positive roots:
this statement is yours and not mine and I wanted to ask why this statement.This statement should not be there. (post 3 your statement)
anemone said:Ah! I see...yes, I re-read post #3 and you are right, I said the given cubic polynomial has exactly three positive roots when it has 2 positive and 1 negative roots, I don't know where my head was when I said that, hehehe...sorry!
I will fix that third post and thanks for catching, Kali!
The formula for finding $b(a+c)$ given roots of a cubic equation is $b(a+c) = -\frac{a^2 - 2ac + c^2}{3}$.
The roots of a cubic equation can be found using the cubic formula $x = \sqrt[3]{\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$.
No, the quadratic formula can only be used to find the roots of a quadratic equation (equations in the form of $ax^2 + bx + c = 0$).
If the cubic equation has complex roots, the formula for finding $b(a+c)$ may still be used, but the result will also be complex.
No, the formula is the most efficient way to find $b(a+c)$ given the roots of a cubic equation. However, it is important to simplify the fraction in the formula to avoid errors in calculations.