"Solving the equatioX2+(3+i)X+1+i=0" was answered by our experts

atenienseec1p

Answered question

2022-04-01

Solving the equatio

X^{2} + (3 + i) X + 1 + i = 0

jmroberts70pbo2

Beginner2022-04-02Added 10 answers

Step 1
You're work so far is good, and indeed you want to find

a, b \in B R

such that

{(a + b i)}^{2} = 4 + 2 i

i.e.

a^{2} - b^{2} = 4

and

2 a b = 2

.
The latter shows you that a and b are nonzero, and that

b = a^{- 1}

. Plugging this into the former yields

a^{2} - a^{- 2} = 4,

,
and multiplying everything by

a^{2}

and rearranging shows that

a^{4} - 4 a^{2} - 1 = 0 .

This is a quadratic in

a^{2}

, so by the quadratic formula

a^{2} = 2 \pm \sqrt{5}

. Because a is real

a = \pm \sqrt{2 + \sqrt{5}} .

Boehm98wy

Beginner2022-04-03Added 18 answers

Step 1

{(a^{2} + b^{2})}^{2} = {(a^{2} - b^{2})}^{2} + 4 a^{2} b^{2}

a^{2} + b^{2} = \sqrt{16 + 4}

and

a^{2} - b^{2} = 4

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