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Solve the following quadratic equation: x2 – (5 – i)x + (18 + i) = 0 - Mathematics and Statistics

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Sum

Solve the following quadratic equation:

x2 – (5 – i)x + (18 + i) = 0

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Solution

Given equation is x2 – (5 – i)x + (18 + i) = 0
Comparing with ax2 + bx + c = 0, we get
a = 1, b = – (5 – i), c = 18 + i
Discriminant = b2 – 4ac
= [–(5 – i)]2 – 4 x 1 x (18 + i)
= 25 – 10i + i2 – 72 – 4i
= 25 – 10i – 1 – 72 – 4i        ...[∵ i2 = – 1]
= – 48 – 14i
So, the given equation has complex roots.
These roots are given by

x = `(-"b" ± sqrt("b"^2 - 4"ac"))/(2"ac")`

= `(-[-(5 - "i")] ± sqrt(-48 - 14"i"))/(2(1)`

= `((5 - "i") ± sqrt(-48 - 14"i"))/2`

Let `sqrt(-48 - 14"i")` = a + bi, where a, b ∈ R
Squaring on both sides, we get
– 48 – 14i = a2 + b2i2 + 2abi
∴ – 48 – 14i = (a2 – b2) + 2abi       ...[∵ i2 = – 1]
Equating real and imaginary parts, we get
a2 – b2 = – 48 and 2ab = – 14

∴ a2 – b2 = – 48 and b = `(-7)/"a"`

∴ `"a"^2 - ((-7)/"a")^2` = – 48

∴ `"a"^2 - 49/"a"^2` = – 48

∴ a4 –  49 = –  48a2
∴ a4 + 48a2 – 49 = 0
∴ (a2 + 49)(a2 – 1) = 0
∴ a2 = – 49 or a2 = 1
But a ∈ R
∴ a2 ≠ – 49
∴ a2 = 1
∴ a = ± 1
When a = 1, b = `-7/1` = –  7

When a = –  1, b = `(-7)/(-1)` = 7

∴ `sqrt(-48 - 14"i")` = ± (1 –  7i)

∴ x = `((5 - "i") ± (1 - 7"i"))/2`

∴ x = `(5 - "i" + 1 - 7"i")/2 or x  = (5 - "i" - 1 + 7"i")/2`

∴ x = 3 – 4I or x = 2 + 3i.

Concept: Solution of a Quadratic Equation in Complex Number System
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APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 11th Standard Maharashtra State Board
Chapter 3 Complex Numbers
Exercise 3.2 | Q 4. (iii) | Page 40
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