If two roots of the equation 4x³ + 20x² - 23x + 6 = 0 are equal, then roots of the equation are –

A repeated root of $f(x) = 4x^3 + 20x^2 - 23x + 6 = 0$ must also satisfy $f'(x) = 12x^2 + 40x - 23 = (2x - 1)(6x + 23) = 0$. Checking $x = \frac{1}{2}$: $f(1/2) = 4(1/8) + 20(1/4) - 23(1/2) + 6 = 0$. Since the sum of the roots is $-\frac{20}{4} = -5$, the third root is $-5 - (1/2 + 1/2) = -6$. Thus the roots are $\frac{1}{2}, \frac{1}{2}, -6$.