If equation $(k - 1) x^{2} + (k^{2} + 1) x + 6 = 0$ and $2 x^{2} + 10 x + 12 = 0$ have both roots common, then the value of $k$ is ___

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2.0

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2.00

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Solution

Since both the roots are common

$\frac{k - 1}{2} (1)$ = $\frac{k^{2} + 1}{10} (2)$ = $\frac{6}{12} (3)$

$\begin{matrix} F r o m (1) a n d (2) & F r o m (2) a n d (3) & F r o m (1) a n d (3) \frac{k - 1}{2} = \frac{k^{2} + 1}{10} & \frac{k^{2} + 1}{10} = \frac{6}{12} & \frac{k - 1}{2} = \frac{6}{12} 5 k - 5 = k^{2} + 1 & k^{2} = 5 - 1 & k = 2 k^{2} - 5 k + 6 = 0 & k^{2} = 4 k = 2, 3 & k = 2, - 2 \end{matrix}$

The value of $k$ is $= 2$ {common value of $k$ is $2$ }

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