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Let $Z_1$ and $Z_2$ be two roots of the equation $x^2+a Z+b=0$ being complex. Further, assume that the origin, $Z_1$ and $\mathrm{Z}_2$ form an equilateral triangle. Then
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The correct answer is:
$a^2=3 b$
$a^2=3 b$
$\mathrm{z}^2+\mathrm{az}+\mathrm{b}=0 ; \mathrm{z}_1+\mathrm{z}_2=-\mathrm{a} \& \mathrm{z}_1 \mathrm{z}_2=\mathrm{b}$
$0, \mathrm{z}, \mathrm{z}_2$ form an equilateral $\Delta$
$\therefore 0^2+\mathrm{z}_1{ }^2+\mathrm{z}_2{ }^2=0 . \mathrm{z}_1+\mathrm{z}_1 \cdot \mathrm{z}_2+\mathrm{z}_2 \cdot 0$
(for equation $\Delta, \mathrm{z}_1{ }^2+\mathrm{z}_2{ }^2+\mathrm{z}_3{ }^2=\mathrm{z}_1 \mathrm{z}_2+\mathrm{z}_2 \mathrm{z}_3+\mathrm{z}_3 \mathrm{z}_1$ ) $z_1^2+z_2^2=z_1 z_2$ or $\left(z_1+z_2\right)^2=3 z_1 z_2$
$\therefore \mathrm{a}^2=3 \mathrm{~b} \text {. }$
$0, \mathrm{z}, \mathrm{z}_2$ form an equilateral $\Delta$
$\therefore 0^2+\mathrm{z}_1{ }^2+\mathrm{z}_2{ }^2=0 . \mathrm{z}_1+\mathrm{z}_1 \cdot \mathrm{z}_2+\mathrm{z}_2 \cdot 0$
(for equation $\Delta, \mathrm{z}_1{ }^2+\mathrm{z}_2{ }^2+\mathrm{z}_3{ }^2=\mathrm{z}_1 \mathrm{z}_2+\mathrm{z}_2 \mathrm{z}_3+\mathrm{z}_3 \mathrm{z}_1$ ) $z_1^2+z_2^2=z_1 z_2$ or $\left(z_1+z_2\right)^2=3 z_1 z_2$
$\therefore \mathrm{a}^2=3 \mathrm{~b} \text {. }$
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