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

Using the sum and product of roots for a^2x^2 + x + (1 - a^2) = 0, we get alpha^2 - beta^2 = -1/a^2 and alpha^2(-beta^2) = (1 - a^2)/a^2. Since alpha^2 = beta^2 - 1/a^2, substituting this into the product equation gives (beta^2 - 1/a^2)(-beta^2) = (1 - a^2)/a^2. Expanding and rearranging results in a^2(beta^2)^2 - (a^2 - 1)beta^2 - 1 = 0. Solving this quadratic equation for beta^2 yields the roots 1 and -1/a^2. Since beta is real, beta^2 must be positive, so beta^2 = 1.

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