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If $x^2+2 x+2 x y+m y-3$ has two rational factors, then the value of $m$ will be
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The correct answer is:
$6,-2$
Given expression $\quad x^2+2 x+2 x y+m y-3$ can be written as $x^2+2 x(1+y)+(m y-3)$
But factors are rational, so $B^2-4 A C$ is a perfect square.
Now $\left.4(1+y)^2-(m y-3)\right\} \geq 0$
$\Rightarrow 4\left\{y^2+1+2 y-m y+3\right\} \geq 0$
$\Rightarrow y^2+2 y-m y+4 \geq 0$
Hence $2 y-m y= \pm 4 y$ {as it is perfect square }
Now taking (-) sign, we get $m=6$
But factors are rational, so $B^2-4 A C$ is a perfect square.
Now $\left.4(1+y)^2-(m y-3)\right\} \geq 0$
$\Rightarrow 4\left\{y^2+1+2 y-m y+3\right\} \geq 0$
$\Rightarrow y^2+2 y-m y+4 \geq 0$
Hence $2 y-m y= \pm 4 y$ {as it is perfect square }
Now taking (-) sign, we get $m=6$
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