Maharashtra State BoardHSC Commerce 11th
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Solve the following quadratic equation: x2-(32+2i)x+62i = 0 - Mathematics and Statistics

Sum

Solve the following quadratic equation:

`x^2 - (3 sqrt(2) + 2"i") x + 6 sqrt(2)"i"` = 0

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Solution

Given equation is `x^2 - (3 sqrt(2) + 2"i") x + 6 sqrt(2)"i"` = 0
Comparing with ax2 + bx + c = 0, we get
a = 1, b = `-(3 sqrt(2) + 2"i"), "c" = 6sqrt(2)"i"`
Discriminant = b2 – 4ac
= `[-(3 sqrt(2) + 2"i")]^2 - 4 xx 1 xx 6 sqrt(2) "i"`

= `18 + 12 sqrt(2) "i" + 4"i"^2 - 24sqrt(2)"i"`

= `18 - 12sqrt(2) "i" - 4`       ...[∵ i2 = – 1]
= `14 - 12sqrt(2)"i"`
So, the given equation has complex roots.
These roots are given by

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

= `(-[-(3sqrt(2) + 2"i")] ± sqrt(14 - 12sqrt(2)"i"))/(2(1)`

= `((3sqrt(2) + 2"i") ± sqrt(14 - 12sqrt(2)"i"))/2`

Let `sqrt(14 - 12sqrt(2)"i")` = a + bi, where a, b ∈ R
Squaring on both sides, we get
`14 - 12sqrt(2) "i"` = a2 + i2b2 + 2abi
∴ `14 - 12sqrt(2) "i"` = (a2 – b2) + 2abi    ...[∵ i2 = – 1]
Equating real and imaginary parts, we get
a2 – b2 = 14 and 2ab = `-12sqrt(2)`

∴ a2 – b2 = 14 and b = `(-6sqrt(2))/"a"`

∴ `"a"^2 - ((-6sqrt(2))/"a")^2` = 14

∴ `"a"^2 - 72/"a"^2` = 14

∴ a4 – 72 = 14a2
∴ a4 –  14a2 –  72 = 0
∴ (a2 –  18) (a2 + 4) = 0
∴ a2 = 18 or a2 = – 4
But a ∈ R
∴ a2 ≠ – 4
∴ a2 = 18
∴ a = ± `3sqrt(2)`
When a = `3sqrt(2), "b" = (-6sqrt(2))/(3sqrt(2)` = – 2

When a = `-3sqrt(2), "b" = (-6sqrt(2))/(-3sqrt(2)` = 2

∴ `sqrt(14 - 12sqrt(2)"i") = ± (3sqrt(2) - 2i")`

∴ x = `((3sqrt(2) + 2"i") ± (3sqrt(2) - 2"i"))/2`

∴ x = `((3sqrt(2) + 2"i") + (3sqrt(2) - 2"i"))/2`

or x = `((3sqrt(2) + 2"i") - (3sqrt(2) - 2"i"))/2`

∴ x = `3sqrt(2)` or x = 2i.

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. (ii) | Page 40
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