Question

x^2-(5-i)x+(18+i)=0 find the quadratic equation

Answer 1

x²-(7-i)x+(18-i)=0

Equation should be:

ax²+bx+c=0,we get a=1,b=-(7-i) & c=18-i

Whereby,α=-b+√b²-4α/2α & β=-b-√b²-4α/2α

Let a+ib=√-24-10i

⇒a²-b²+2abi=-24-10i

⇒a²-b²=-24 & 2ab=-10

consider:

{a²+b²}²=(a²-b²)²+4a²b²

⇒{a²+b²)²=(-24)^4+(-10)²=676

⇒a²+b²=26

Therefore,a²-b²=-24 &a²+b²=26

⇒a²=1 & b²=25

⇒a=+/-1 & b=+/-5

But 2ab=-10,so a & b must be opposite signs:

Therefore α=1 &b=-5 or α=-1 & b=5

⇒√-24-100=1-5i /-1+5i

Substituting either of these values in equation (i),we get

α=(7-i)+(1-5i)/2 & β=(7-i)-(1-5i)/2

⇒α=4-3i & β=3-3i

Therefore, roots of the given equation are 4-3i & 3-3i