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+1 vote
2.2k views
in Polynomials by (20 points)
edited by

If the Zeroes of the Quadratic Polynomial 6x²- 3 - 7x are α and β then find the quadratic polynomial whose zeroes are α/β and β /α .

the options are

a) 5x2 -6x + 4

b) 7x+ x + 1

c) 18x+ 85x + 1

d)15x- 5x + 3

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1 Answer

+2 votes
by (40.3k points)

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

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