Question

If $\alpha$ is a complex, such that $\alpha$$z^2$ + z +$\overline{\alpha }$ = 0 has a real root. Then

• A. + = 1
• B. + = -1
• C. Both A and B
• D. None of these

Answer 1

The correct option is C

Both A and B

$\alpha$$z^2$ + z +$\overline{\alpha }$ = 0 --------------(1)

Let $\alpha$ = x + iy

$\overline{\alpha }$ = x - iy

(x + iy)$z^2$ + z + (x - iy) = 0

Let the real root be p

Substitute the p in place of z

(x + iy)$p^2$ + p + (x - iy) = 0-------------(2)

(x$p^2$ + p + x) + (y$p^2$ - y)i = 0

Equating real and imaginary parts on both sides

x$p^2$ + px + 1 = 0 or y$p^2$ - 2 = 0

$p^2$ = 1.

p = $\underset{–}{+}1$

Substituting p value in equation 2

$\alpha$$(\underset{–}{+}1{)}^2$ + $\left(\underset{–}{+}1\right)$ + $\overline{\alpha }$ = 0

$\alpha$ + $\overline{\alpha }$ = $\underset{–}{+}1$.