# Equation problem

1. Nov 21, 2009

### 1/2"

Hey there!!
I have a few doubts regarding this worked out example in my book,
It goes like this…
Solve 2a ^ (3/2) +(a^2-x^2) ^ (1/2) {(a+x) ^ (1/2) +(a-x) ^ (1/2) }= a^ (2/3) +a{(a+x) ^(1/2)+(a-x) ^1/2}+(a^2-x^2) ^ (1/2)
Well now they say
Hence, removing from a^ ( 3/2) both sides and transposing, we get
a^ (1/2) {a-(a^2-x^2) ^ (1/2) }= {(a+x) ^ (1/2) + (a-x ) ^ (1/2) }x { a-(a^ 2 -x^2) ^ (1/2) }
whence a^ (1/2) = ( a+x) ^ (1/2) +(a-x) ^ (1/2) ;
squaring both sides a= 2a =2( a^2-x^2) ^ (1/2)
anyway it concludes as x= ± ( a√ 3)/2
But the problem is that I can’t understand how could 2a^2/3 suddenly disappear!!
Plz help me!!

Last edited: Nov 22, 2009
2. Nov 21, 2009

### tiny-tim

Hey 1/2"!

(try using the X2 tag just above the Reply box )
It's very difficult to read your equations, but I think the answer is that 2a2/3 hasn't disappeared …

a2/3 has disappeared from both sides, which is ok because you can always add or subtract the same amount from both sides (or indeed multiply or divide both sides by the same amount, or raise them to the same power)

3. Nov 22, 2009

### 1/2"

i am sorry for the mistakes !!:tongue:
I have corrected them . For futher mistake please let post it!!

4. Nov 22, 2009

### tiny-tim

Hi 1/2"!

(just got up :zzz: …)

First, you don't need ^ if you're using SUP (maybe in handwriting, where different levels aren't clear, but certainly not in typing), and you don't need brackets round (1/2) etc.

Quicker if I do that now …
ok, now I think I can see what's happened …

either you or your book are misprinting "2/3" for "3/2"

the a1/2 {a that begins the second equation-line is the same as a3/2, which is presumably what it should be throughout.

Does that sort it out?

5. Nov 23, 2009

### 1/2"

I'm sorry i don't get it.

6. Nov 23, 2009

### tiny-tim

Well, the a2/3 just after the = sign doesn't seem to match anything else in that equation, so I think it's a misprint for a3/2