Find the value of $k$ for which quadratic equation $(k - 2) x^{2} + 2 (2 k - 3) x + 5 k - 6 = 0$ has equal roots

k = 3

1

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k = - 2

- 1

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k = - 3

1

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k = 2

2

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Solution

The correct option is A $k = 3$ or $1$
Given expression $(k - 2) x^{2} + 2 (2 k - 3) x + 5 k - 6 = 0$ has equal roots.
Therefore, $D = b^{2} - 4 a c = 0$
We have $a = (k - 2), b = 2 (2 k - 3), c = 5 k - 6$
Thus ${(2 (2 k - 3))}^{2} - 4 (k - 2) (5 k - 6) = 0$
$\Rightarrow [{(2 k - 3)}^{2} - (k - 2) (5 k - 6)] = 0$
$\Rightarrow [(4 k^{2} + 9 - 12 k) - (5 k^{2} - 16 k + 12)] = 0$
$\Rightarrow [4 k^{2} + 9 - 12 k - 5 k^{2} + 16 k - 12] = 0$
$\Rightarrow [- k^{2} + 4 k - 3] = 0$
$\Rightarrow k^{2} - 4 k + 3 = 0$
$\Rightarrow (k - 1) (k - 3) = 0$
$\Rightarrow$ $k = 1$ or $k = 3$

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