Question

If $a$ and $b$ are real and $3+i$ is a root of $x^2+ax+b=0$, then $a$ is

• A. $3$
• B. $-3$
• C. $6$
• D. $-6$

Answer 1

Answer: (D)

$-6$

Since the coefficients are real, hence the roots will occur as conjugate pairs.
Therefore if one root is $3+i$ then the other root will be $3-i$.
Thus sum of root will be $-a$.
$=3+i+3-i$
$=6$
Hence, $-a=6$ or
$a=-6$ .
Similarly product of roots $=b$
$=(3+i)(3-i)$
$=9+1$
$=10$
Hence, $a=-6$ and $b=10$
Thus the equation changes to
$x^2-6x+10=0$