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How do I solve ${(x - 2)}^{2} = - 3$ $(x-2)^2 = -3$ using the square root property?

1 Answer

Aug 21, 2015

$x = 2 \pm (\sqrt{3}) i$ $x= 2+-(sqrt(3))i$

Explanation:

If $a^{2} = b$ $a^2 = b$
then $a = \pm \sqrt{b}$ $a = +-sqrt(b)$

In this case $a = (x - 2)$ $a = (x-2)$ and $b = - 3$ $b = -3$
So
$XXXX x - 2 = \pm \sqrt{- 3}$ $color(white)("XXXX")x-2 = +-sqrt(-3)$

Note that $\sqrt{- 3} = \sqrt{3} \cdot \sqrt{- 1}$ $sqrt(-3) = sqrt(3)*sqrt(-1)$
and there is no Real value $= \sqrt{- 1}$ $= sqrt(-1)$ .
However, if we are allowed to use Complex values
by definition $i = \sqrt{- 1}$ $i =sqrt(-1)$

Therefore, we can write
$XXXX x - 2 = \pm (\sqrt{3}) i$ $color(white)("XXXX")x-2 =+- (sqrt(3))i$
and after adding 2 to both sides
$XXXX x = 2 \pm (\sqrt{3}) i$ $color(white)("XXXX")x = 2+-(sqrt(3))i$

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