Question:
The equation of the smallest degree with real coefficients having 1 + i as one of the roots is
(a) $x^{2}+x+1=0$
(b) $x^{2}-2 x+2=0$
(c) $x^{2}+2 x+2=0$
(d) $x^{2}+2 x-2=0$
Solution:
(b) $x^{2}-2 x+2=0$
We know that, imaginary roots of a quadratic equation occur in conjugate pair.
It is given that, $1+i$ is one of the roots.
So, the other root will be $1-i$.
Thus, the quadratic equation having roots $1+i$ and $1-i$ is,
$x^{2}-(1+i+1-i) x+(1+i)(1-i)=0$
$\Rightarrow x^{2}-2 x+2=0$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.