Option (C) is correct 18x2 + 85x + 18 = 0
Given that α and β are zeros of polynomial 6x2 - 7x - 3
\(\therefore\) α + β = \(\cfrac{-b}a=\cfrac{-(-7)}6=\cfrac76\)
And αβ = \(\cfrac{c}a=\cfrac{-3}6=\cfrac12\)
Now, (α + β)2 = \(\cfrac{49}{36}\)
⇒ α2 + β2+ 2αβ = \(\cfrac{49}{36}\)
Now, \(\cfrac{\alpha^2+\beta^2+2\alpha \beta}{\alpha\beta}=\cfrac{\frac{49}{36}}{\frac{-1}{2}}=\cfrac{49}{18}\)
⇒ \(\cfrac{\alpha}{\beta}+\cfrac{\beta}{\alpha}+2=\cfrac{49}{18}\)
⇒ \(\cfrac{\alpha}{\beta}+\cfrac{\beta}{\alpha}=\cfrac{-49}{18}-2\)
= \(\cfrac{-49-36}{18}\)
= \(\cfrac{-85}{18}\)
And \(\cfrac{\alpha}{\beta}\times\cfrac{\beta}{\alpha}=1\)
Therefore, the polynomial whose zeroes are \(\cfrac{\alpha}{\beta}\) and \(\cfrac{\beta}{\alpha}\) is,
x2 - \(\left(\cfrac{\alpha}{\beta}+\cfrac{\beta}{\alpha}\right)\)x + \(\cfrac{\alpha}{\beta}+\cfrac{\beta}{\alpha}=0\)
⇒ x2 - \(\left(-\cfrac{85}{18}\right)\)x + 1 = 0
⇒ 18x2 + 85x + 18 = 0
Hence, required quadratic polynomial in 18x2 + 85x + 18 = 0
Hence, option (c) is correct