x^2 (5 i)x+18+i=0 Substitue x=a+ib (a+bi)^2 (5 i)(a+bi)+18+i=0 (a^2 b^2 5a b+18)+i(1+a 5b+2ab)=0 [ a^2 b^2 5a b+18=0; 1+a 5b+2ab=0 ...

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Algebra of Complex Numbers

Solve x2-5-i...

Question

Solve
$x^{2} - (5 - i) x + (18 + i) = 0$

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Solution

$x^{2} - (5 - i) x + 18 + i = 0$
Substitue $x = a + i b$

${(a + b i)}^{2} - (5 - i) (a + b i) + 18 + i = 0$

$(a^{2} - b^{2} - 5 a - b + 18) + i (1 + a - 5 b + 2 a b) = 0$

$[\begin{matrix} a^{2} - b^{2} - 5 a - b + 18 = 0 1 + a - 5 b + 2 a b = 0 \end{matrix}]$

solving the above equations, we get,

$(\begin{matrix} a = 2, & b = 3 a = 3, & b = - 4 \end{matrix})$

$x = 2 + 3 i, 3 - 4 i$

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