Solve the following equation:
$x^{2} - 7 x + 12 = 0$

x_{1} = 1, x_{2} = 2

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x + 1 = 2, x + 2 = 3

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x_{1} = 3, x_{2} = 4

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x_{1} = 5, x_{2} = 6

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Solution

The correct option is B $x_{1} = 3, x_{2} = 4$
$x^{2} - 7 x + 12 = 0$

$\Rightarrow x^{2} - 3 x - 4 x + 12 = 0$

$\Rightarrow x (x - 3) - 4 (x - 3) = 0$

$\Rightarrow (x - 3) (x - 4) = 0$

$x - 3 = 0$ $x - 4 = 0$

$x = 3$ $x = 4$

$∴ x_{1} = 3; x_{2} = 4$

Hence, correct option is $x_{1} = 3, x_{2} = 4$ .

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