$y = x^{2} + x - 10$
$y = 3 x - 2$
The equations above intersect each other at two points.Which of the following is true about both points of intersection?

x > - 3

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x > - 2

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y < 4

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y > 5

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Solution

The correct option is B $x > - 2$
$y = x^{2} + x - 10$
$\Rightarrow y = 3 x - 2$
on comparing
$\Rightarrow x^{2} + x - 10 = 3 x - 2$
$\Rightarrow x^{2} + x - 3 x - 10 + 2 = 0$
$\Rightarrow x^{2} - 2 x - 8 = 0$
$\Rightarrow x^{2} - 4 x + 2 x - 8 = 0$
$\Rightarrow x (x - 4) + 2 (x - 4) = 0$
$\Rightarrow (x - 4) (x + 2) = 0$
$\Rightarrow x = - 2$ $,$ $x = 4$
$∴ x > - 2$
Hence, the answer is $x > - 2.$

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