Question

# If $$\alpha$$ and $$\beta$$ use the zeroes of the polynomial $$x^2 + px + q$$, then find the value of $$\left[\dfrac{\alpha}{\beta} + 2\right] \times \left[\dfrac{\beta}{\alpha} + 2 \right]$$.

Solution

## Given equation: $$x^2+px+q$$So, $$a=1, b=p$$ and $$c=q$$$$\left [ \dfrac{\alpha }{\alpha }+2 \right ]\left [ \dfrac{\beta }{\alpha }+2 \right ]$$$$=1+2\left ( \dfrac{\alpha ^2+\beta ^2}{\alpha \beta } \right )+4$$$$=2\left ( \dfrac{\left (\alpha +\beta \right )^2-2\alpha \beta}{\alpha \beta } \right )+5$$$$\alpha +\beta=\dfrac{-b}{a}, \alpha \beta=\dfrac{c}{a}$$$$=2\left ( \dfrac{\left (\dfrac{-b}{a} \right )^2-2\dfrac{c}{a}}{\dfrac{c}{a}} \right )+5$$$$=2\left ( \dfrac{\dfrac{b^2-2ac}{a^2}}{\dfrac{c}{a}} \right )+5$$$$=2\left ( {\dfrac{b^2-2ac}{ac}} \right )+5$$$$=2\left ( {\dfrac{b^2}{ac}}-2 \right )+5$$Since, $$a=1, b=p$$ and $$c=q$$$$\Rightarrow \left [ \dfrac{\alpha }{\alpha }+2 \right ]\left [ \dfrac{\beta }{\alpha }+2 \right ]=2\left ( {\dfrac{p^2}{q}}-2 \right )+5$$Mathematics

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