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If $\alpha,\beta $ are real and $z$ is complex number and $z^2+\alpha z +\beta=0$ has two distinct roots on the line $R_e z=1,$ then the values of $\beta$=?

(A) $\beta\in (0,1)$ (B) $\beta\in (-1,0)$ (C) $\beta\in (1,\infty)$ (D) $\beta\in (-1,1)$

1 Answer

Toolbox:
  • If the roots of $ax^2+bx+c=0$ are distinct and complex, then Discriminent $D<0$
  • $D=b^2-4ac$
Given: $z^2+\alpha z+\beta=0$has rwo distinct roots on $R_ez=1$
and $z$ is complex
Roots of the eqn., $z=\large\frac{-\alpha \pm\sqrt{\alpha^2-4\beta}}{2}$
since z is complex, $D=\alpha^2-4\beta<0$
Also given that $R_ez=1$, $\large-\frac{\alpha}{2}=1$
$\Rightarrow\:\alpha=-2$
$\therefore\:4-4\beta<0$
$\beta>1$
$\Rightarrow\:\beta\in (1,\infty)$
answered Jul 18, 2013 by rvidyagovindarajan_1
 

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