Find the value of

Question

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)

Bunuel · Accepted Answer

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.

mbaiseasy · Answer

Remember these things regarding absolute values in the GMAT (1) \sqrt{x^2}=|x| (2) |x| = x ==> if x > 0 (3) |x| = -x ==> if x < 0 Golden rules! Must memorize! Solution: ***Transform the equation: \frac{|1+a| + |a-1|}{|1+a| - |a-1|} ***Since we know that a is a positive fraction as given: 0

sharmila79 · Answer

Hi Bunuel,
Could you please explain me how a-1 becomes 1-a in the numerator and denominator in the second part of the problem?
Thanks!

sanjoo · Answer

Bunuel ..i ddnt get it.. how did u get the 1/a in the end ?? can u plz explain bunuel..? m trying it but i cant get 1/a in the end?

Bunuel · Answer

\frac{|1+a|+|a-1|}{|1+a|-|a-1|}=\frac{(1+a)+(1-a)}{(1+a)-(1-a)}=\frac{1+a+1-a}{1+a-1+a}=\frac{2}{2a}=\frac{1}{a} .

Hope it's clear now.

Bunuel · Answer

Check this: math-absolute-value-modulus-86462.html Absolute value properties: When x\leq{0} then |x|=-x , or more generally when some \ expression\leq{0} then |some \ expression|={-(some \ expression)} . For example: |-5|=5=-(-5) ; When x\geq{0} then |x|=x , or more generally when some \ expression\geq{0} then |some \ expression|={some \ expression} . For example: |5|=5 ; So, for our case, since a<1 , then a-1<0 hence according to the above |a-1|=-(a-1)=1-a . Hope it helps.

karjan07 · Answer

Bunuel,

When to use 
 
1.\sqrt{x} = Mod(X) or

2. \sqrt{X} = X

Here u have used \sqrt{x} = Mod(X) while in the problem below:

If \sqrt{3-2x}= \sqrt{2x} +1 then 4x2 =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

you have used \sqrt{X} = X

Why so?

Bunuel · Answer

\sqrt{x^2}=|x| . 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 . I guess you are talking about the following problem: if-root-3-2x-root-2x-1-then-4x-135539.html Where did I write that \sqrt{x^2}=x ? P.S. Please read this: rules-for-posting-please-read-this-before-posting-133935.html#p1096628 Press m button when formatting. Thank you.

karjan07 · Answer

Got it... My mistake... was confused on  \sqrt{(3x-2)} ^2 =  3-2x

Probably should sleep now !!

Bunuel · Answer

Right. In that question we have (\sqrt{3-2x})^2 , which equals to 3-2x , the same way as (\sqrt{x})^2=x . If it were \sqrt{(3-2x)^2} , then it would equal to |3-2x| , the same way as \sqrt{x^2}=|x| . Check here: if-rot-3-2x-root-2x-1-then-4x-107925.html#p1223681 Hope it's clear.

WholeLottaLove · Answer

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 · Answer

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 .
___________________________________

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.

mumbijoh · Answer

Hi Bunuel,
I was also thinking along the same lines as you in this question.However, I got stuck deciding which rule to apply to which equation.
How did you decide this. Couldn't you have swapped the rules then?
|1+a|=-1-a
And |a-1|=a-1

Bunuel · Answer

Absolute value properties: 
When  x\leq{0}  then  |x|=-x , or more generally when  some \ expression\leq{0}  then  |some \ expression|={-(some \ expression)} . For example:  |-5|=5=-(-5) ;

When  x\geq{0}  then  |x|=x , or more generally when  some \ expression\geq{0}  then  |some \ expression|={some \ expression} . For example:  |5|=5 ;

So, for our case, since  a<1 , then  a-1<0  hence according to the above  |a-1|=-(a-1)=1-a .

Similarly as  0<a<1 , then  1+a>0 , hence  |1+a|=1+a .

Hope it helps.

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