Solve: x2-(3√2-2i)x-√2i=0.
Solution of a quadratic equation by Shridhracharya rule:
Given quadratic equation is
x2-(3√2-2i)x-√2i=0
Here, a=1,b=-32-2i,c=-2i
Substituting a and b and c in
We know:
x=-b±b2-4ac2a
⇒x=32-2i±32-2i2+42i2.........equation1
Let 32-2i2+42i =x+yi
⇒32-2i2+42i=x+yi2⇒14-82i=x2-y2+2xyi⇒x2-y2=142xy=-82............equation2
x2+y22=x2-y22+4x2y2⇒x2+y22=196+128=324⇒x2+y2=18.......equation3
From eq.2 and eq.3,
⇒x=-4,y=2or x=4,y=-2
⇒x+iy=4-2i,or -4+2i
⇒x+iy=±4-2i
Substituting these equations in eq.1⇒x=32-2i±4-2i2.
Solve the following quadratic equations :
(i)x2−(3√2+2i)x+6√2i=0
(ii)x2−(5−i)x+(18+i)=0
(iii)(2+i)x2−(5−i)x+2(1−i)=0
(iv)x2−(2+i)x−(1−7i)=0
(v)ix2−4x−4i=0
(vi)x2+4ix−4=0
(vii)2x2+√15ix−i=0
(viii)x2−x+(1+i)=0
(ix)ix2−x+12i=0
(x)x2−(3√2−2i)x−√2i=0
(xi)x2−(√2+i)x+√2i=0
(xii)2x2−(3+7i)x+(9i−3)=0
Solve the following quadratic equations by factorization method :
(i) x2+10ix−21=0
(ii) x2+(1−2i)x−2i=0
(iii) x2−(2√3+3i)x+6√3i=0
(iv)6x2−17ix−12=0