Question

If one of the roots of the equation $2x^2-6x+k=0$ is $(\alpha +5i)/2$ , then the values of $\alpha$ and $k$ are

• A. $\alpha =3,k=8$
• B. $\alpha =\frac{3}{2},k=17$
• C. $\alpha =2,k=16$
• D. $\alpha =3,k=\frac{17}{2}$

Answer 1

Answer: (D)

$\alpha =3,k=\frac{17}{2}$

complex roots always occur in conjugate pair

$2x^2-6x+k=0$

$sum=3$

product = $\frac{k}{2}$

one root is $\frac{\alpha +5i}{2}$

other root is $\frac{\alpha -5i}{2}$

$sum=\frac{\alpha +5i}{2}+\frac{\alpha -5i}{2}=3$

$\alpha =3$

product = $k$ = $\left(\frac{3+5i}{2}\right)\left(\frac{3-5i}{2}\right)$

$=\frac{9-25i^2}{4}=\frac{9+25}{4}=\frac{17}{2}$