Question

If $\alpha$ and $\beta$ are the roots of the equation ${ax}^2+bx+c=0$ ($a\ne 0,a.b,c$ being different), then $(1+\alpha +{\alpha }^2)(1+\beta +{\beta }^2)$ is equal to

• A. Zero
• B. Positive
• C. Negative
• D. None of these

Answer 1

Answer: (A), (B)

The correct options are
A Zero
B Positive

$Given$
$a{x}^2+bx+c=0\alpha ,\beta aretheroots$
$\alpha +\beta =\frac{-b}{a},\alpha \beta =\frac{c}{a}$
$(1+\alpha +{\alpha }^2)(1+\beta +{\beta }^2)$

$=1+\alpha +\beta +{\alpha }^2+{\beta }^2+\alpha \beta (\alpha +\beta )+\alpha \beta +{\left(\alpha \beta \right)}^2$
$=1+\alpha +\beta +{(\alpha +\beta )}^2-2\alpha \beta +\alpha \beta (\alpha +\beta )+\alpha \beta +{\left(\alpha \beta \right)}^2$
$=1-\frac{b}{a}+{\left(\frac{-b}{a}\right)}^2-2\frac{c}{a}+\frac{c}{a}\left(\frac{-b}{a}\right)+\frac{c}{a}+{\left(\frac{c}{a}\right)}^2$
$=1-\frac{b}{a}+\frac{b^2}{a^2}-\frac{2c}{a}-\frac{cb}{a^2}+\frac{c}{a}+\frac{c^2}{a^2}$
$=1-\frac{b}{a}+\frac{b^2}{a^2}-\frac{c}{a}-\frac{cb}{a^2}+\frac{c^2}{a^2}$
$=\frac{a^2-ab+b^2-ac-cb+c^2}{a^2}$
$=\frac{a^2+b^2+c^2-ab-ac-cb}{a^2}$
$mulitplyanddivideby2$
$=\frac{2a^2+2b^2+2c^2-2ab-2ac-2cb}{2a^2}$
$=\frac{a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ac}{2a^2}\frac{}{}$
$=\frac{{(a-b)}^2+{(b-c)}^2+{(c-a)}^2}{2a^2}\ge 0$
$=positiveor0$