What must be added to $x^{2} + 3 x - 8$ to get $x^{2} + x + 6$ ?

3 - 8

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- 2x + 14

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x + 6

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2x + 14

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Solution

The correct option is B
- 2x + 14

Let the algebraic expression that is to be added be $P$ .

Then,

$(x^{2} + 3 x - 8) + P = x^{2} + x + 6$

$\Rightarrow P = (x^{2} + x + 6) - (x^{2} + 3 x - 8)$

$\Rightarrow P = x^{2} + x + 6 - x^{2} - 3 x + 8$
$\Rightarrow P = - 2 x + 14$
So, $- 2 x + 14$ must be added to $x^{2} + 3 x - 8$ to get $x^{2} + x + 6$ .

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