Question

If $\alpha ,\beta ,\gamma$ are the roots of $x^3+2x^2-3x+1=0$, then

• A. $\frac{\alpha \beta }{\alpha +\beta }+\frac{\alpha \gamma }{\alpha +\gamma }+\frac{\beta \gamma }{\beta +\gamma }=\frac{11}{7}$
• B. $\frac{\alpha \beta }{\alpha +\beta }+\frac{\alpha \gamma }{\alpha +\gamma }+\frac{\beta \gamma }{\beta +\gamma }=-\frac{11}{7}$
• C. $\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=3$
• D. $\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=-3$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{\alpha \beta }{\alpha +\beta }+\frac{\alpha \gamma }{\alpha +\gamma }+\frac{\beta \gamma }{\beta +\gamma }=\frac{11}{7}$
C $\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=3$

Given equation is $x^3+2x^2-3x+1=0$ has roots as $\alpha ,\beta ,\gamma$
Now, assuming
$S=\frac{\alpha \beta }{\alpha +\beta }+\frac{\alpha \gamma }{\alpha +\gamma }+\frac{\beta \gamma }{\beta +\gamma }\Rightarrow S=\frac{1}{\frac{1}{\alpha }+\frac{1}{\beta }}+\frac{1}{\frac{1}{\alpha }+\frac{1}{\gamma }}+\frac{1}{\frac{1}{\beta }+\frac{1}{\gamma }}$
Finding the equation whose roots are $\frac{1}{\alpha },\frac{1}{\beta },\frac{1}{\gamma }$
$x^3-3x^2+2x+1=0$
Here, $\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=3$,
So, $S=\frac{1}{3-\frac{1}{\gamma }}+\frac{1}{3-\frac{1}{\beta }}+\frac{1}{3-\frac{1}{\alpha }}$
$\Rightarrow S=-[\frac{1}{\frac{1}{\gamma }-3}+\frac{1}{\frac{1}{\beta }-3}+\frac{1}{\frac{1}{\alpha }-3}]$

Finding the equation whose roots are $\frac{1}{\alpha }-3,\frac{1}{\beta }-3,\frac{1}{\gamma }-3$
$(x+3{)}^2-3(x+3{)}^2+2(x+3)+1=0\Rightarrow x^3+6x^2+11x+7=0$

Finding the equation whose roots are $\frac{1}{\frac{1}{\alpha }-3},\frac{1}{\frac{1}{\beta }-3},\frac{1}{\frac{1}{\gamma }-3}$
$7x^3+11x^2+6x+1=0$

Therefore, $S=-(-\frac{11}{7})$
$\Rightarrow S=\frac{11}{7}$