What is the equation whose roots are α and β, if \((\frac{-α^2}{β})\) and \((\frac{-β^2}{α})\) are roots of the equation \(3x^2 – 18x + 2 = 0\)?
A. \(3x^2 + 6x + 2 = 0\)
B. \(3x^2 - 6x + 2 = 0\)
C. \(3x^2 + 6x - 2 = 0\)
D. \(3x^2 - 6x - 2 = 0\)
E. \(3x^2 - 6x - 3 = 0\)
I believe there is an error somewhere. The math ain't mathing. I could be misguided in my math here.
For simplicity, I will refer alpha as 'a' and beta as 'b'.
From equation 2:
a) Product:
\(\frac{2}{3} = (\frac{-a^2}{b})(\frac{-b^2}{a})\)
\(\frac{2}{3} = -a*-b\)
\(2 = 3ab\)
\(4 = 6ab\) ----a
b) Sum:
\(\frac{18}{3} = (\frac{-a^2}{b}) + (\frac{-b^2}{a})\)
\(6 = \frac{(-a^3 - b^3)}{ab}\)
\(6ab = -a^3 - b^3\) (substitute the value of 6ab)
\(4 = -a^3 -b^3\) ----b
The Rationale:
From eq a, for 6ab to result in positive number, either
both a & b must be negative or both positive.
Now eq b helps us narrow down further:
either both a^3 and b^3 must be negative to give a positive sum,
OR the larger one of them should be negative.
So using eq a and b together, we get to know that both a (alpha) and b (beta) must be negative.
However, the answer to this question requires both to be positive (as the sum and product are both positive in the equation). Which goes against the equation we have derived.
I could be wrong here.
Bunuel please help us.