If $\alpha$ and $\beta$ are the roots of the equation $x^2-3x-1=0$, then form a quadratic equation whose roots are$\dfrac{1}{\alpha^2} $ and $\dfrac{1}{\beta^2}$.

For the equation x^2-3x-1=0, the sum of roots a+b = 3 and the product ab = -1. The new roots are 1/a^2 and 1/b^2, so the new sum is (a^2+b^2)/(a^2*b^2) = ((a+b)^2-2ab)/(ab)^2 = (3^2-2(-1))/(-1)^2 = 11/1 = 11. The new product is 1/(a^2*b^2) = 1/(-1)^2 = 1. The required equation is x^2-11x+1=0.

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