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Combination

If x + 12 =...

Question

If $(x + 1)^{2} = 4$ and $(x - 1)^{2} = 16$ , find the value of $x$ .

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Solution

The correct option is A $- 3$
As, $(x + 1)^{2} = 4$
It can be written as, $x^{2} + 2 x + 1 = 4$ .......(1)
and $(x - 1)^{2} = 16$
$∴ x^{2} - 2 x + 1 = 16$ ......(2)
Substituting (2) from (1), we get
$x^{2} + 2 x + 1 - (x^{2} - 2 x + 1) = 4 - 16$
$\Rightarrow 4 x = - 12$
$\Rightarrow x$ $= - 3$
Hence, option A is correct.

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