Question

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

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

Answer 1

Answer: (B)

$1$

Since $({\alpha }^2,-{\beta }^2)$ are the roots of the given equation, we have

${\alpha }^2-{\beta }^2=-\frac{1}{a^2}.......\left(1\right)$ and

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

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

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

Substituting for $-\frac{1}{a^2}$ from $\left(1\right)$ into $\left(2\right)$, we have

${\alpha }^2{\beta }^2-1={\alpha }^2{-\beta }^2$

$\Rightarrow {\alpha }^2{\beta }^2-{\alpha }^2-1+{\beta }^2=0$ $\Rightarrow ({\alpha }^2+1)(1-{\beta }^2)=0$

$\Rightarrow 1-{\beta }^2=0$

$\Rightarrow {\beta }^2=1$

Hence, ${\beta }^2=1$