Solve the following quadratic equations:
x2 – (2 + i)x – (1 – 7i) = 0
x2 – (2 + i)x – (1 – 7i) = 0
Given x2 – (2 + i)x – (1 – 7i) = 0
Recall that the roots of quadratic equation ax2 + bx + c = 0, where a ≠ 0, are given by
Here, a = 1, b = –(2 + i) and c = –(1 – 7i)
By substituting i2 = –1 in the above equation, we get
We can write 7 – 24i = 16 – 9 – 24i
⇒ 7 – 24i = 16 + 9(–1) – 24i
⇒ 7 – 24i = 16 + 9i2 – 24i [∵ i2 = –1]
⇒ 7 – 24i = 42 + (3i)2 – 2(4)(3i)
⇒ 7 – 24i = (4 – 3i)2 [∵ (a – b)2 = a2 – b2 + 2ab]
By using the result 7 – 24i = (4 – 3i)2, we get
∴ x = 3 – i or –1 + 2i
Thus, the roots of the given equation are 3 – i and –1 + 2i.