Q. 2 B4.1( 7 Votes )

# Solve the following quadratic equations:x2 – (5 – i)x + (18 + i) = 0

x2 – (5 – i)x + (18 + i) = 0

Given x2 – (5 – i)x + (18 + i) = 0

Recall that the roots of quadratic equation ax2 + bx + c = 0, where a ≠ 0, are given by

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

By substituting i2 = –1 in the above equation, we get

By substituting –1 = i2 in the above equation, we get

We can write 48 + 14i = 49 – 1 + 14i

48 + 14i = 49 + i2 + 14i [ i2 = –1]

48 + 14i = 72 + i2 + 2(7)(i)

48 + 14i = (7 + i)2 [ (a + b)2 = a2 + b2 + 2ab]

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

[ i2 = –1]

x = 2 + 3i or 3 – 4i

Thus, the roots of the given equation are 2 + 3i and 3 – 4i.

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