Square Root Method

**~berserk** · October 25th 2009, 01:58 PM

I have the problem: $(x+1)^2=3$

the answers are: $-1-\sqrt{3}, -1+\sqrt{3}$

I need to know how to arrive at the answer

**skeeter** · October 25th 2009, 02:04 PM

Originally Posted by ~berserk

I have the problem: $(x+1)^2=3$

the answers are: $-1-\sqrt{3}, 1-\sqrt{3}$

I need to know how to arrive at the answer

one of your "answers" is incorrect ...

$(x+1)^2 = 3$

$x+1 = \pm \sqrt{3}$

$x = -1 \pm \sqrt{3}$

**~berserk** · October 25th 2009, 02:09 PM

Corrected, thank you

**e^(i*pi)** · October 25th 2009, 02:14 PM

Originally Posted by ~berserk

I have the problem: $(x+1)^2=3$

the answers are: $-1-\sqrt{3}, -1+\sqrt{3}$

I need to know how to arrive at the answer

I shan't go through the answer as skeeter has done that (and it's the best method for solving the equation when it's set up as in the question)

As $(a)^2 = (-a)^2 \: , \: a \in \mathbb{R}$ we cannot be sure if we are getting the negative solution or the positive one. In fact both are equally valid.
If it is easier you can use the difference of two squares to show this

$a^2 = b^2 \: \rightarrow \: a = b$ - this is not entirely true:

If $a^2=b^2$ then it follows that $a^2-b^2=0$ . By the difference of two squares:

$a^2-b^2 = (a-b)(a+b)=0 \: \therefore a = \pm b$

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