Question

If $(3+i)$ is a root of the equation $x^2+ax+b=0$, where $a$ and $b$ are rationals, then $a$ is equal to

• A. $3$
• B. $-3$
• C. $6$
• D. $-6$

Answer 1

The correct option is D $-6$

Since all the coefficients are natural, hence the complex roots occur in conjugate pairs.
Thus if one root is $(3+i)$ then the conjugate root will be $(3-i)$.
Thus sum of roots is
$=3+i+(3-i)$
$=6$ ...(i)
And product of roots will be
$=(3+i)(3-i)$
$=10$...(ii)
Now consider a quadratic equation of the form
$A{x}^2+Bx+C=0$
Then,
Sum of roots $=\frac{-B}{A}$.
Product of roots $=\frac{C}{A}$.
In the above case
$\frac{-B}{A}=6$ and $\frac{C}{A}=10$.
Thus the equation becomes
$x^2+\frac{B}{A}x+\frac{C}{A}=0$
$x^2-6x+10=0$
Comparing with the given equation $x^2+ax+b=0$
Hence we obtain, $a=-6$ and $b=10$.