isi-entrance 2012 Q30

isi-entrance · India · solved Roots of polynomials Vieta's formulas: compute symmetric functions of roots

Let $s, sr, sr^2, sr^3$ be the roots of $x^4 + ax^3 + bx^2 + cx + d = 0$ (roots in geometric progression). Show that $c^2 = a^2 d$.

With roots $s, sr, sr^2, sr^3$: $-a = s(r^4-1)/(r-1)$, $c = s^3r^3(r^4-1)/(r-1)$, $d = s^4r^6$. Then $c^2 = s^6r^6\{(r^4-1)/(r-1)\}^2 = (s^4r^6)[s(r^4-1)/(r-1)]^2 = d \cdot a^2$. Hence $c^2 = a^2d$. (D)

Let $s, sr, sr^2, sr^3$ be the roots of $x^4 + ax^3 + bx^2 + cx + d = 0$ (roots in geometric progression). Show that $c^2 = a^2 d$.

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