Square rooting an equation/inequality, which side is +/-?

[autodidude]

When you have something like x^2 > 2/3 and you root it, why does the left hand side become the plus or minus and not the right side? I only get the correct answer if I do it on the left hand side.

 

[micromass]

I think the usual way of doing this is not writing +/- in front of anything, but just expanding this equations in two equations.

Let's take an x such that

x^2>4

(I pick 4 because the notation will be easier). When I square root both sides, then I'll get

\sqrt{x^2}>2

However, note the very important point that \sqrt{x^2} is NOT always equal to x. This is only true if x\geq 0. For example: \sqrt{(-1)^2}=1\neq -1.

So we have to split up in two cases:
1) Either x\geq 0. In that case \sqrt{x^2}=x, so we get

x>2

2) Or x<0, in that case \sqrt{x^2}=-x, so we get

-x&gt;2

Or equivalently

x&lt;-2

So, to conclude: if we take an x such that x^2&gt;4, then we either get that x>2 or x<-2.

 

[Mark44]

To expand slightly on what micromass said, while it is not true that
$\sqrt{x^2} = x$
it is true that $\sqrt{x^2} = |x|$.

Getting rid of the absolute values leads to the same two cases that micromass mentioned.

 

[autodidude]

^ Thanks, but don't understand why you can't put a plus/minus (besides from the fact that you get an incorrect answer :p) in front of the number

 

[Mark44]

Because, by definition, the square root of a nonnegative real number is nonnegative.

For example, many people erroneously believe that √4 = ±2. Although 4 does have two square roots, the principal square root of 4 is 2.

 

[HallsofIvy]

When you have something like x^2 > 2/3 and you root it, why does the left hand side become the plus or minus and not the right side? I only get the correct answer if I do it on the left hand side.

It doesn't. If x^2&gt; 2/3 then either x&lt;-\sqrt{2/3} or x&gt; \sqrt{2/3}.

Conversely, if the problem were x^2&lt; 2/3 then -\sqrt{2/3}&lt; x&lt; \sqrt{2/3}.

 

[autodidude]

^ Thanks HallsOfIvy, never knew that!

 

[DonAntonio]

Because, by definition, the square root of a nonnegative real number is nonnegative.

For example, many people erroneously believe that √4 = ±2. Although 4 does have two square roots, the principal square root of 4 is 2.

Well, it is NOT erroneous to "believe" that \sqrt{4}=\pm 2 since, as it happens, both values on the RHS when

squared equal 4 and this is the primary definition of "square root.

It is DEFINED that \sqrt{4}=2 mostly, I think, to make \sqrt{x} a function, which otherwise it wouldn't be. If one want to mess with the

negative root is thus customary to take -\sqrt{4}=-2 and everybody happy.

DonAntonio

 

[Mark44]

Well, it is NOT erroneous to "believe" that \sqrt{4}=\pm 2 since, as it happens, both values on the RHS when

squared equal 4 and this is the primary definition of "square root.

Granted, 4 has two square roots, and I mentioned this earlier in the thread. However, the notation $\sqrt{a}$ indicates the principal square root.

It is DEFINED that \sqrt{4}=2 mostly, I think, to make \sqrt{x} a function, which otherwise it wouldn't be. If one want to mess with the

negative root is thus customary to take -\sqrt{4}=-2 and everybody happy.

DonAntonio

 

[autodidude]

Ooh, one more thing, does this apply to trig equations as well?

e.g. cos^2(x) = 1/4 becomes cos(x)=1/2 and -cos(x)=1/2 (from ±√(cos^2(x)) = √(1/2)) rather than cos(x)=±1/2 (which evaluates to the same answer, just being pedantic)

 

[Mark44]

Ooh, one more thing, does this apply to trig equations as well?



e.g. cos^2(x) = 1/4 becomes cos(x)=1/2 and -cos(x)=1/2

Yes, of course. The above should be cos(x) = 1/2 OR -cos(x) = 1/2.

(from ±√(cos^2(x)) = √(1/2)) rather than cos(x)=±1/2 (which evaluates to the same answer, just being pedantic)

 

[autodidude]

^ Thank you! Ah, missed that >.<