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Relationship Between Zeroes and Coefficients of a Quadratic Polynomial

If α and ...

Question

If $α$ and $β$ are the zeroes of the polynomial $x^{2} - 3 x - 2$ , find the quadratic polynomial whose zeroes are $\frac{1}{2 α + β}$ and $\frac{1}{2 β + α}$ .

15 x^{2} - 9 x + 2

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16 x^{2} + 9 x - 1

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16 x^{2} - 9 x + 1

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Solution

The correct option is C $16 x^{2} - 9 x + 1$
Given that $α$ & $β$ are zero of polynomial
$f (x) = x^{2} - 3 x - 2$
therefore $α + β = 3$
$α β = - 2$
Now, the zero of the required quadratic polynomial are,
$\frac{1}{2 α + β}$ & $\frac{1}{2 β + α}$
Sum of the roots-
$\frac{1}{2 α + β} + \frac{1}{2 β + α} = \frac{2 β + α + 2 α + β}{(2 α + β) (2 β + α)} = \frac{3 (α + β)}{4 α β + 2 α^{2} + 2 β^{2} + α β}$
$= \frac{3 \times 3}{4 \times (- 2) + 2 [(α + β)^{2} - 2 α β] + (- 2)}$
$= \frac{9}{- 10 + 2 [9 + 2 \times 2]}$
$= \frac{9}{- 10 + 26}$
$= \frac{9}{16}$
Products of roots:-
$\frac{1}{2 α + β} \times \frac{1}{2 β + α} = \frac{1}{4 α β + 2 [(α + β)^{2} - 2 α β] + α β} = \frac{1}{16}$
Now Req eq.
$x^{2} - ($ sum of roots $)$ x $+$ Product of roots $= 0$
$= x^{2} - \frac{9}{16} x + \frac{1}{16} = 0$
$= 16 x^{2} - 9 x + 16 = 0$ .

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