If $x = 3 + i$ then prove that $x^{2} - 6 x + 13 = 0$ .

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Solution

Given,
$x = 3 + 2 i$
or, $x - 3 = 2 i$
Now squaring both sides we get,
$(x - 3)^{2} = - 4$
or, $x^{2} - 6 x + 9 = - 4$
or, $x^{2} - 6 x + 13 = 0$ .

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