WholeLottaLove wrote:
Hello
I am having a tough time figuring out why |a-1| = -(a-1)
I know that for some absolute value problems, you test the positive and negative cases of a number. For example, 4=|2x+3| I would solve for:
I. 4=2x+3
II. 4=-(2x+3)
But I am not sure why you are deriving |a-1| = -(a-1) here. Does it have something to do with a being positive? (i.e. 0<a<1)
Also, in another problem I just solved, I ended up with |5-x| = |x-5| which ended up being |5-x| = |5-x| Why wouldn't |a+1| = |a-1|
Thanks as always!
Bunuel wrote:
manimani wrote:
Find the value of \(\frac{\sqrt{(1+a)^2} + \sqrt{(a-1)^2}}{\sqrt{(1+a)^2} - \sqrt{(a-1)^2}}\) for 0<a<1
A. a
B. 1/a
C. (a-1)/(a+1)
D. (a+1)/(a-1)
E. a/(a-1)
Note that \(\sqrt{x^2}=|x|\).
\(\frac{\sqrt{(1+a)^2} + \sqrt{(a-1)^2}}{\sqrt{(1+a)^2} - \sqrt{(a-1)^2}}=\frac{|1+a|+|a-1|}{|1+a|-|a-1|}\)
Now, since \(0<a<1\), then: \(|1+a|=1+a\) and \(|a-1|=-(a-1)=1-a\).
Hence \(\frac{|1+a|+|a-1|}{|1+a|-|a-1|}=\frac{(1+a)+(1-a)}{(1+a)-(1-a)}=\frac{1}{a}\).
Answer: B.
Hope it's clear.
If \(x\leq{0}\). then \(\sqrt{x^2}=|x|=-x\). For example if \(x=-5\), then \(\sqrt{(-5)^2}=\sqrt{25}=5=|x|=-x\).
If \(x\geq{0}\), then \(\sqrt{x^2}=|x|=x\). For example if \(x=5\), then \(\sqrt{5^2}=\sqrt{25}=5=|x|=x\).
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We are given that \(0<a<1\).
For this range \(1+a>0\), so \(|1+a|=|positive|=1+a\)
For this range \(a-1<0\), so \(|a-1|=|negative|=-(a-1)=1-a\).
Hope it's clear.