A quadratic equation with real coefficients whose one root is $3 - 2 i$ is

x^{2} - 3 x + 2 = 0

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x^{2} - 6 x + 13 = 0

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x^{2} - 2 x + 3 = 0

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None of these

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Solution

The correct option is A $x^{2} - 6 x + 13 = 0$
Complex roots occur in conjugate form.
Hence,
Conjugate of $3 - 2 i$ will be $3 + 2 i$ .
Thus, the roots are $3 - 2 i$ and $3 + 2 i$ .
Sum of the roots, is $6$ .
And product of the roots is $(3 - 2 i) (3 + 2 i)$ $= 9 + 4$ $= 13$ .
Hence, the equation will be
$x^{2} - 6 x + 13 = 0$ .

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(3 + 2 \sqrt{3})

x^{2} + 6 x - 3 = 0

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x^{2} + 6 x + 3 = 0

x^{2} - 6 x + 3 = 0

x^{2} - 3 x + 2 = 0.

A = {x \in R : x^{2} - | x | - 2 = 0}

B = {α + β, α β}

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x^{2} + | x | - 2 = 0 .

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x^{2} - 2 x + 3 = 0

2

f (x) = (\begin{matrix} x^{2} - 1, if 0 < x < 2 2 x + 3, if 2 \leq x < 3 \end{matrix}

lim x \to 2^{-} f (x) and lim x \to 2^{+} f (x)

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