Question

If $\alpha ,\beta$ are roots of $p_1{x}^2+q_{1}x+r_1=0,\beta ,\gamma$ are roots of $p_2{x}^2+q_{2}x+r_2=0$ and $\alpha ,\beta$ are the roots of $p_3{x}^2+q_{3}x+r_3=0$ then

• A. $\alpha +\beta +\gamma =\frac{-1}{2}\left[\frac{q_1{p}_2{p}_3+q_2{p}_1{p}_3+p_1{p}_2{q}_3}{p_1{p}_2{p}_3}\right]$
• B. $\alpha +\beta +\gamma =\frac{-1}{2}\left[\frac{p_1{q}_2{q}_3+p_2{q}_1{p}_3+q_1{q}_2{p}_3}{p_1{p}_2{p}_3}\right]$
• C. $\prod _{i=1}^3\left(p_i+r_i-q_i\right)=p_1{p}_2{p}_2\left(1+\sum \alpha +\sum \alpha \beta +\sum \alpha \beta \gamma \right)$
• D. $\prod _{i=1}^3\left(p_i+r_i-q_i\right)=p_1{p}_2{p}_2\left(1-\sum \alpha +\sum \alpha \beta -\sum \alpha \beta \gamma \right)$

Answer 1

The correct option is A $\alpha +\beta +\gamma =\frac{-1}{2}\left[\frac{q_1{p}_2{p}_3+q_2{p}_1{p}_3+p_1{p}_2{q}_3}{p_1{p}_2{p}_3}\right]$

$p_1{x}^2+q_{1}x+r_1=0$

$p_2{x}^2+q_{2}x+r_2=0$

$p_3{x}^2+q_{3}x+r_3=0$

$x+p=\frac{q_1}{p_1}---\left(1\right)$

$p+y=\frac{q_1}{p_2}---\left(2\right)$

$x+y=\frac{q_3}{p_3}---\left(3\right)$

Adding (1), (2) and (3) we get

$\alpha +\beta +\gamma =\frac{-1}{2}\left[\frac{q_1{p}_2{p}_3+q_2{p}_1{p}_3+p_1{p}_2{q}_3}{p_1{p}_2{p}_3}\right]$