Question

If $\alpha ,\beta$ are real and ${\alpha }^2,-{\beta }^2$ are the roots of $a^2{x}^2+x+1-a^2=0;(a>1)$, then ${\beta }^2=$

• A. $a^2$
• B. $1$
• C. $1-a^2$
• D. $1+a^2$

Answer 1

The correct option is B $1$

${\alpha }^2,-{\beta }^2$ are the roots of $a^2{x}^2+x+1-a^2=0$
$\Rightarrow {\alpha }^2-{\beta }^2=-\frac{1}{a^2},$

And

$\Rightarrow {\alpha }^2(-{\beta }^2)=\frac{1-a^2}{a^2}$

$({\alpha }^2+{\beta }^2{)}^2=({\alpha }^2-{\beta }^2{)}^2+4{\alpha }^2{\beta }^2$

$=\frac{1}{a^4}+4\left(\frac{a^2-1}{a^2}\right)=\frac{1}{a^4}+4-\frac{4}{a^2}$

$={(2-\frac{1}{a^2})}^2\Rightarrow {\alpha }^2+{\beta }^2=2-\frac{1}{a^2}$

${\beta }^2=\frac{1}{2}\left[\right({\alpha }^2+{\beta }^2)-({\alpha }^2-{\beta }^2\left)\right]$

$=\frac{1}{2}[2-\frac{1}{a^2}+\frac{1}{a^2}]=1$