# Algebra - general

• Mar 9th 2009, 03:30 AM
sammy28
Algebra - general
hello to all.

this happens to be a surd but could equally apply to any algebraic equation involving fractions.

ive been asked to solve the following in the form $\displaystyle A\sqrt{B}+C$, where A, B and C are rational numbers.

$\displaystyle \frac{1}{\sqrt{2}}+x = \frac{x}{\sqrt{2}}$

my question is given any algebraic fraction should you always begin by multiplying the denominator?

because i get two answers depending on which way i go but only one is in the requested format, and thats the one that multiplies the denominator to start with, eg:

$\displaystyle \frac{\sqrt{2}}{\sqrt{2}} + \sqrt{2}x = \frac{\sqrt{2}x}{\sqrt{2}}$
$\displaystyle 1 = x - x\sqrt{2}$
$\displaystyle 1 = x(1 - \sqrt{2})$
$\displaystyle x = -1-\sqrt{2}$

the other way give the correct answer if checked on a calculator but is not in the correct format

$\displaystyle \frac{1}{\sqrt2} = \frac{x}{\sqrt{2}} - x$
$\displaystyle \frac{1}{\sqrt{2}} = x(\frac{1}{\sqrt{2}}-1)$
$\displaystyle \frac{1}{1-\sqrt{2}}$

you could rationlize this surd which then give the same answer as above but this seems a bit long winded.
• Mar 9th 2009, 03:40 AM
Quote:

Originally Posted by sammy28
hello to all.

this happens to be a surd but could equally apply to any algebraic equation involving fractions.

ive been asked to solve the following in the form $\displaystyle A\sqrt{B}+C$, where A, B and C are rational numbers.

$\displaystyle \frac{1}{\sqrt{2}}+x = \frac{x}{\sqrt{2}}$

my question is given any algebraic fraction should you always begin by multiplying the denominator?

because i get two answers depending on which way i go but only one is in the requested format, and thats the one that multiplies the denominator to start with, eg:

$\displaystyle \frac{\sqrt{2}}{\sqrt{2}} + \sqrt{2}x = \frac{\sqrt{2}x}{\sqrt{2}}$
$\displaystyle 1 = x - x\sqrt{2}$
$\displaystyle 1 = x(1 - \sqrt{2})$
----------------
ERROR HERE
$\displaystyle x = -1-\sqrt{2}$
----------------------------
correct one

$\displaystyle 1 = x(1 - \sqrt{2})$

Divide both sides by $\displaystyle (1 - \sqrt{2})$

$\displaystyle x= 1/(1 - \sqrt{2})$

See quote
• Mar 9th 2009, 05:27 AM
sammy28
thanks ardash

i dont think thats an error, if you rationalize the denominator you get

$\displaystyle \frac{1}{1-\sqrt{2}} * \frac{1+\sqrt{2}}{1+\sqrt{2}}$

$\displaystyle \frac{1 + \sqrt{2}}{-1}$

$\displaystyle -1-\sqrt{2}$

my question was really is it always quicker to factor out the denominator first or does it really not matter
• Mar 9th 2009, 06:15 AM
repcvt
Quote:

Originally Posted by sammy28
hello to all.

this happens to be a surd but could equally apply to any algebraic equation involving fractions.

ive been asked to solve the following in the form $\displaystyle A\sqrt{B}+C$, where A, B and C are rational numbers.

$\displaystyle \frac{1}{\sqrt{2}}+x = \frac{x}{\sqrt{2}}$

my question is given any algebraic fraction should you always begin by multiplying the denominator?

because i get two answers depending on which way i go but only one is in the requested format, and thats the one that multiplies the denominator to start with, eg:

$\displaystyle \frac{\sqrt{2}}{\sqrt{2}} + \sqrt{2}x = \frac{\sqrt{2}x}{\sqrt{2}}$
$\displaystyle 1 = x - x\sqrt{2}$
$\displaystyle 1 = x(1 - \sqrt{2})$
$\displaystyle x = -1-\sqrt{2}$

the other way give the correct answer if checked on a calculator but is not in the correct format

$\displaystyle \frac{1}{\sqrt2} = \frac{x}{\sqrt{2}} - x$
$\displaystyle \frac{1}{\sqrt{2}} = x(\frac{1}{\sqrt{2}}-1)$
$\displaystyle \frac{1}{1-\sqrt{2}}$

you could rationlize this surd which then give the same answer as above but this seems a bit long winded.

I think got your point ,the question can be see why
• Mar 9th 2009, 07:54 AM
Quote:

Originally Posted by sammy28
thanks ardash

i dont think thats an error, if you rationalize the denominator you get

$\displaystyle \frac{1}{1-\sqrt{2}} * \frac{1+\sqrt{2}}{1+\sqrt{2}}$

$\displaystyle \frac{1 + \sqrt{2}}{-1}$

$\displaystyle -1-\sqrt{2}$

my question was really is it always quicker to factor out the denominator first or does it really not matter

Sorry, I never tried to rationalize it ,
And no it is not always quicker to factor out the denominator first but most of the times your speed will completely depend on the amount of your practice of any method or a "particular" question