Question

If $\alpha$ is an unreal constant such that $\alpha z^2+z+\overline{\alpha }=0$ has a real root, then

• A. $\alpha +\overline{\alpha }=\pm 1$
• B. $\alpha +\overline{\alpha }=0$
• C. $\alpha +\overline{\alpha }=-2$
• D. None of these

Answer 1

Answer: (A)

$\alpha +\overline{\alpha }=\pm 1$

Let a=x+iya=x+iy where x,y∈Rx,y∈R

Then the equation becomes

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

$=(x{z}^2+z+x)+i(y{z}^2-y)$

0+0i=0=(x+iy)z2+z+(x−iy)

=(xz2+z+x)+i(yz2−y)

Since $(x{z}^2+z+x)$and $(y{z}^2-y)$are real, we have a real solution to our equation if the equations

$x{z}^2+z+x=0$ and $y(z+1)(z-1)$ have a common real solution.

If y=0y=0 and x=0x=0, we have z=0z=0 as a real solution. In this case 2x=02x=0

Otherwise if y=0y=0 and x≠0x≠0,

Every real number zz satisfies y(z+1)(z−1)=0y(z+1)(z−1)=0

Thus, the original equation has a real solution iff the equation

$x{z}^2+z+x=0$ has a real solution.

$0=x{z}^2+z+x=x(z+12x)2-14x+x$

Or $(z+\frac{1}{2}x)2=\frac{1}{4x^2}-1$

xz2+z+x=0

0=xz2+z+x=x(z+12x)2−14x+x

(z+12x)2=14x2−1

Thus, we have a real solution iff

$\dfrac{1}{4x^{2}}-1 >= 0$

or, $x^2$ ≤ $(\frac{1}{2}{)}^2$

or, $xϵ[\frac{-1}{2},\frac{1}{2}]$-{0}

Therefore, $2xϵ[-1,1]$-{0}

14x2−1≥0

x2≤(12)2

x∈[−12,12]−{0}

2x∈[−1,1]−{0}

If y≠0y≠0

z=1z=1 or z=−1z=−1

If z=1z=1 is a solution, then x+1+x=0x+1+x=0, or 2x=−12x=−1

If z=−1z=−1 is a solution, then x−1+x=0x−1+x=0, or 2x=12x=1

Therefore, the original equation has a real solution iff

a+a¯=2Re(a)=2x∈[−1,1]