Question

x2 – (5 – i) x + (18 + i) = 0
​

Answer 1

Answer:

x2 – (5 – i) x + (18 + i) = 0

Given as x^2 – (5 – i) x + (18 + i) = 0

Now we shall apply discriminant rule,

Where, x = (-b ±√(b^2 – 4ac))/2a

Here, a = 1, b = -(5 - i), c = (18 + i)

Therefore,

Then, we have i^2 = -1

On substituting -1 = i^2 in the above equation, we get

Now, we can write as 48 + 14i = 49 – 1 + 14i

Therefore,

48 + 14i = 49 + i2 + 14i [∵ i^2 = –1]

= 72 + i2 + 2(7)(i)

= (7 + i)^2 [Since, (a + b)^2 = a^2 + b^2 + 2ab]

On using the result 48 + 14i = (7 + i)^2, we get

 x = 2 + 3i or 3 – 4i

Therefore, the roots of the given equation are 3 – 4i, 2 + 3i

Answer 2

Answer:

your answer is in the above picture