If α, β are the zeros of the polynomial x2 + 6x + 2, then

Question:

If $\alpha, \beta$ are the zeros of the polynomial $x^{2}+6 x+2$, then $\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=?$

(a) 3
(b) −3
(c) 12
(d) −12

 

Solution:

(b) $-3$

Since $\alpha$ and $\beta$ are the zeroes of $x^{2}+6 x+2$, we have:

$\alpha+\beta=-6$ and $\alpha \beta=2$

$\therefore\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=\left(\frac{\alpha+\beta}{\alpha \beta}\right)=\frac{-6}{2}=-3$

 

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