If $\displaystyle a\in R, b\in R$, then the equation $\displaystyle x^{2}-abx-a^{2}=0$ has

To determine the nature of the roots, we evaluate the discriminant of the quadratic equation, which is $D = b^2 - 4ac$. Substituting the coefficients gives $(-ab)^2 - 4(1)(-a^2)$, simplifying to $a^2b^2 + 4a^2$, or $a^2(b^2 + 4)$. Because the square of any real number is non-negative, $D$ is strictly greater than zero, meaning the roots are real and distinct. Additionally, the product of the roots is $c/a = -a^2$, which is always negative, proving there is one positive root and one negative root.

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