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If the lines represented by $\left(\mathrm{k}^2+2\right) \mathrm{x}^2+3 \mathrm{xy}-6 \mathrm{y}^2=0$ are perpendicular to each other, then the values of $\mathrm{K}$ are
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The correct answer is:
$\pm 2$
The lines $\left(k^2+2\right) x^2+3 x y-6 y^2=0$ are perpendicular to each other.
$\therefore\left(\mathrm{k}^2+2\right)+(-6)=0 \Rightarrow \mathrm{k}^2=4 \Rightarrow \mathrm{k}= \pm 2$
$\therefore\left(\mathrm{k}^2+2\right)+(-6)=0 \Rightarrow \mathrm{k}^2=4 \Rightarrow \mathrm{k}= \pm 2$
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