Algebra question

  • Thread starter spynjr
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  • #1
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Hi all

I've been stumped on a basic question.

Show that:

[tex] \frac{{\sqrt x - \sqrt a }}{{x - a}} = \frac{1}{{\sqrt x + \sqrt a }} [/tex]

I've tried removing the radical sign, bringing the denominator up, etc...

Help?
 

Answers and Replies

  • #2
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I'll post the answer below, but I strongly advise you not look and give it more shots unless your really low on time.

Hint: multiply by some form of one.









Answer: multiply by x^1/2 -a^1/2/x^1/2 -a^1/2 (which is one)
 
  • #3
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Thanks for that. I did try that but couldn't get sqrt(x) + sqrt(a) as the denominator.

I can see that it's correct numerically though.

Is that a general algebraic rule or can you get to it algebraically?
 
  • #4
symbolipoint
Homework Helper
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BIG HINT: Multiply the right-side by 1 (look back to post #2).

Maybe you could not read through the text straight-line-text form of the expression. See as
[tex] \[
\frac{{\sqrt x - \sqrt a }}{{\sqrt x - \sqrt a }}
\]
[/tex]
 
  • #5
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Hmm, I see that, but don't know how you get to sqrt(x)+sqrt(a)

Nonetheless I've figured out the way that makes most sense to me. Simply factor x-a which equals (sqrt(x)-sqrt(a))(sqrt(x)+sqrt(a)) (difference of two squares) then cancel (sqrt(x)-sqrt(a))
 
  • #6
cristo
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Hmm, I see that, but don't know how you get to sqrt(x)+sqrt(a)

Nonetheless I've figured out the way that makes most sense to me. Simply factor x-a which equals (sqrt(x)-sqrt(a))(sqrt(x)+sqrt(a)) (difference of two squares) then cancel (sqrt(x)-sqrt(a))
Well done: that's the way I would do it!
 
  • #7
445
3
I'll post the answer below, but I strongly advise you not look and give it more shots unless your really low on time.

Hint: multiply by some form of one.









Answer: multiply by x^1/2 -a^1/2/x^1/2 -a^1/2 (which is one)
Woops, change those minus to plus haha.

Unlessn you apply it to the RH =P
Conjugates.
 
  • #8
Brother to equal R.H.S with L.H.S plz follow the pic given in attachment.
 

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  • #9
epenguin
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In case you think "I can follow the calculation but how ever did they think of doing it that way?" the more usual problem set (in exercises for the purpose, but also in real mathematical necessity is to do it the other way round, to get from your right hand side to the left. To get rid of surds in a denominator. You have probably done exercises like that so realise the connection. You will also meet or may already have a similar trick for getting rid of imaginaries in a denominator.

'Difference of two squares' is maybe a refrain to remember.
 

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