a quadratic equation?

Question

please help me solve this

(1+i)r^2-r+(1-i)=0

Y_Y i feel so dumb
thanks in advance!!

Answer 1

So, it's already in Ax^2 +Bx+C=0 where A=(1+i), x=r, b=1, and c=(1-i).
So, plug it into the Quadratic Fromula and solve for x(r in this case)
x=(-b +/- sqrt (b^2-4ac))/(2ab)
r=(-1 +/- sqrt (1^2-4(1+i)(1-i)))/(2(1+i)(1))
r= -1 +/- sqrt (1-(-4i^2)+8i+4)/(2i+2)
So now split it into two equations, one with + and one with -.
r= (-1+1+4i+sqrt8i+2)/(2i+2)
Again split it into 2 more equations, one dividing by 2i, the other dividing by 2.
r=[(-1+1+4i+sqrt8i+2)/(2i)] + [(-1+1+4i+sqrt8i+2)/2]
r=3i+2+ [(-1+1+4i+sqrt8i+2)/2]
r=3i+2+ [2i+sqrt4i+1]
r=3i+2+sqrt4i+1
r=3i+sqrt4i+3
or
r=[(-1-1+4i+sqrt8i+2)/(2i)] + [(-1-1+4i+sqrt8i+2)/(2)]
r= -i+2+2+i +[(-1-1+4i+sqrt8i+2/(2)]
r= 4+ [2i+sqrt4i]
r=4+2i+sqrt4i
So finally in conclusion, we have that:

r=3i+sqrt4i+3 OR 2i+sqrt4i+4

Answer 2

=^.^= Don't feel dumb.

For this you'll have to use the quadratic formula.

1 +- rt. [1 - 4 (1 + i)(1 - i)] all over 2(1 + i)

So solve it all out (remember i^2 = -1) and i got

1 +- i rt.7 all over 2 + 2i.

Hope that helps!