Question

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

• A. $5$
• B. $7$
• C. $9$
• D. $\frac{9}{2}$

Answer 1

The correct option is A $5$

$x^3-2x^2+3x-1=0$
Replace $x$ by $\frac{1}{x}$, we have
$\frac{1}{x^3}-\frac{2}{x^2}+\frac{3}{x}-1=0$
$\Rightarrow x^3-3x^2+2x-1=0$

$\alpha +\beta +\gamma =3,\sum \alpha \beta =2$
$\therefore {\alpha }^2+{\beta }^2+{\gamma }^2=(\alpha +\beta +\gamma {)}^2-2\sum \alpha \beta$
$=3^2-2\times 2$
$=5$

Alternate solution:
$x^3-2x^2+3x-1=0$
$\alpha +\beta +\gamma =2$
$\alpha \beta +\beta \gamma +\gamma \alpha =3\alpha \beta \gamma =1$
$\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}=\frac{{\beta }^2{\gamma }^2+{\alpha }^2{\gamma }^2+{\beta }^2{\alpha }^2}{{\alpha }^2{\beta }^2{\gamma }^2}$
$=\frac{(\alpha \beta +\beta \gamma +\gamma \alpha {)}^2-2\sum {\alpha }^2\beta \gamma }{{\alpha }^2{\beta }^2{\gamma }^2}...\left(1\right)$
$\sum {\alpha }^2\beta \gamma =(\alpha +\beta +\gamma )\alpha \beta \gamma =2$
From eqn(1)
$\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}=\frac{3^2-2\times 2}{1^2}=5$