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 . ${{x}^{2}}-2x+2=0$
c . ${{x}^{2}}+2x+2=0$
d . ${{x}^{2}}+2x-2=0$

Answer 1

Hint: In this question, we have to find the equation of the smallest degree having $1+i $ as one of its roots. We know if one root is complex, then the other root is also complex. Then we find the sum and the product of roots and after solving it, we get the quadratic equation.

Formula used:
${{x}^{2}}$- ( sum of roots a and b )$x$+ ( product of roots a and b ) = 0

Complete step by step Solution:
We know standard form of quadratic equation is $a{{x}^{2}}+bx+c=0$
Given one of the complex roots of the equation is $(1+i)$
So, there should be another complex root for the equation as its complement is $(1-i)$ for its coefficients to be rational.
Therefore, the equation of the lowest degree will be the one that has only these two roots.
We know any quadratic equation whose roots are a and b can be formed by writing as
${{x}^{2}}$- ( sum of roots a and b )$x$+ ( product of roots a and b ) = 0
Quadratic equations whose roots are a and b are given as
${{x}^{2}}-(a+b)x+ab=0$
So we will construct a quadratic equation having roots as $(1+i)$and $(1-i)$
Sum of roots = $(1+i)$+ $(1-i)$
                        = 2
Product of roots = $(1+i)$$(1-i)$
                               = $1-{{i}^{2}}$
We know ${{i}^{2}}$= $-1$
So $1-{{i}^{2}}$= $1$$1$$-(-1)$ = 2
So the quadratic equation be ${{x}^{2}}-2x+2=0$

Therefore, the correct option is (b).

Note: It must be remembered that if one of the roots is complex then another root will also be complex. Students must know the complex values to solve these types of questions.