solve the equation 7-(2+i)x=1-7i
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Answer:
Given equation is x2 – (2 + i) x – (1 – 7i) = 0
Comparing with ax2 + bx + c = 0, we get
a = 1, b = – (2 + i), c = –(1 – 7i)
Discriminant = b2 – 4ac
= [– (2 + i)]2 – 4 x 1 – (1 – 7i)
= 4 + 4i + i2 + 4 – 28i
= 4 + 4i – 1 + 4 – 28i ...[∵ i2 = – 1]
= 7 – 24i
So, the given equation has complex roots.
These roots are given by
x =
-b±b2-4ac2a
=
-[-(2+i)]±7-24i2(1)
=
(2+i)±7-24i2
Let 7-24i = a + bi, where a, b ∈ R
Squaring on both sides, we get
7 – 24i = a2 + i2b2 + 2abi
∴ 7 – 24i = (a2 – b2) + 2abi ...[∵ i2 = – 1]
Equating real and imaginary parts, we get
a2 – b2 = 7 and 2ab = – 24
∴ a2 – b2 = 7 and b =
-12a
∴
a2-(-12a)2 = 7
∴
a2-144a2 = 7
∴ a4 – 144 = 7a2
∴ a4 – 7a2 – 144 = 0
∴ (a2 – 16) (a2 + 9) = 0
∴ a2 = 16 or a2 = – 9
But a ∈ R
∴ a2 ≠ – 9
∴ a2 = 16
∴ a = ± 4
When a = 4, b =
-124 = – 3
When a = – 4, b =
-12-4 = 3
∴ 7-24i = ± (4 – 3i)
∴ x =
(2+i)±(4-3i)2
∴ x =
(2+i)+(4-3i)2 orx=(2+i)-(4-3i)2
∴ x = 3 – i or x = – 1 + 2i.
Concept: Solution of a Quadratic Equation in Complex Number System
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Chapter 3: Complex Numbers -
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