Number of values of $k$ so that the equations $x^{2} + k x + (k + 2) = 0 and x^{2} + (1 - k) x + 3 - k = 0$ have exactly one common root, is -

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Solution

The correct option is A $0$
Condition of common root in
$(c a_{1} - a c_{1})^{2} = (b c_{1} - c b_{1}) (a b_{1} - b a_{1})$
$((k + 2) - (3 - k))^{2} = (k (3 - k) - (k + 2) (1 - k)) ((1 - k) . (1) - k))$
$\Rightarrow (2 k - 1)^{2} = 0$
$k = \frac{1}{2}$

But after putting $k$ , we get same equation, so there can not be exactly one common root.

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x^{2} + k x + (k + 2) = 0 and x^{2} + (1 - k) x + 3 - k = 0

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x^{2} + k x + (k + 2) = 0 and x^{2} + (1 - k) x + 3 - k = 0

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