Solve Using the Square Root Method

**fcabanski** · December 11th 2010, 06:59 PM

$(2x-1)^{2}=-4$

I thought this could only be used if k (the constant) was positive.

**dwsmith** · December 11th 2010, 07:02 PM

$\displaystyle \sqrt{(2x-1)^2}=\sqrt{-4}\Rightarrow 2x-1=\pm 2\mathbf{i}\Rightarrow x=\frac{1\pm 2\mathbf{i}}{2}$

**Prove It** · December 11th 2010, 07:03 PM

You are right, this does not have any real solutions.

It does have two complex solutions though...

$\displaystyle (2x - 1)^2 = -4$

$\displaystyle 2x - 1 = \pm \sqrt{-4}$

$\displaystyle 2x - 1 = \pm 2i$

$\displaystyle 2x = 1 \pm 2i$

$\displaystyle x = \frac{1}{2} \pm i$ .

**fcabanski** · December 11th 2010, 07:42 PM

Yes, that's what I came up with, just wasn't certain it was correct.

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