Question

Solve :- x 2 -(7 - i) x + (18 - i) = 0 over C. ANURAG.

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:
Thereforeα=1 &b=-5 orα=-1 & b=5
⇒√-24-100=1-5i /-1+5i
Substituting either of these values in(i),we get
α=(7-i)+(1-5i)/2 & β=(7-i)-(1-5i)/2
⇒α=4-3i & β=3-3i
Hence,rtoots of the given equation are 4-3i & 3-3i

Answer 2

x2 - (7 - i) x + (18 - i) = 0
On comparing the given equation with general equation ax2 + bx + c = 0, we get a = 1, b = - (7 - i) and c = 18 - i

We shall first find:
under root  - 24 - 10i

Let a + ib = under root - 24 -10i

a2 - b2 + 2abi = - 24 - 10i

a2 - b2 = -24 and 2ab = -10

consider , (a2 + b2)2 = (a2 - b2)2 + 4a2b2

(a2 + b2)2 = ( -24)4 + ( - 10)2 = 676

a2 + b2 = 26

Now, a2 - b2 = -24 and a2 + b2 = 26

a2 = 1 and b2 = 25

a = +- 1 and b = +- 5

But 2ab = - 10, so a and b must be of opposite signs.

Therefore, a = 1 and b = - 5 or a = -1 and b = 5 

under root - 24 - 100 = 1 - 5i or -1 + 5i

Substituting either of the values in (i), we get

alpha = (7 - i) + (1 - 5i) / 2 and beta = (7 - i) - (1 - 5i) / 2

alpha = 4 - 3i and beta = 3 - 3i

Hence, roots of the given equation are 4 - 3i and 3 - 3i