The real part of imaginary roots of the equation $x^{3}+x^{2}+x-2=0$ lies in the interval.

Let the three roots of the equation x³ + x² + x - 2 = 0 be the real number r and the complex pair a ± bi. By Vieta's formulas for cubic equations, the sum of the roots is r + 2a = -1. Using trial values in the equation shows that x = 1 is not a root, but testing the interval boundaries reveals the real root lies between 0 and 1. Assuming the real root r is 1, we get 2a = -2, so a = -1. This means the real part of the imaginary roots lies in the interval (-1, -1/2).

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