## Help with Solving an Equation Containing Radicals...

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maroonblazer
Posts: 51
Joined: Thu Aug 12, 2010 11:16 am
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### Help with Solving an Equation Containing Radicals...

Hi,
I've got the following problem:
$y\sqrt{3} + 1 = 3 - y$

I add y to both sides to rid y on the right. Then I subtract 1 from both sides to get this:

$2y\sqrt{3}=2$

Divide both sides by 2 to get:
$y\sqrt{3} = 1$

then divide by $\sqrt{3}$ to get:
$y = \frac{1}{\sqrt{3}}$

multiply the fraction by $\frac{\sqrt{3}}{\sqrt{3}}$ to get rid of the radical and that leaves me with:
$y = \frac{\sqrt{3}}{3}$

The book gives the answer as:
$\sqrt{3} - 1$

I can't tell where I've taken a wrong step.

Ideas?

Thanks in advance,
mb

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
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### Re: Help with Solving an Equation Containing Radicals...

Hi,
I've got the following problem:
$y\sqrt{3} + 1 = 3 - y$

I add y to both sides to rid y on the right. Then I subtract 1 from both sides to get this:

$2y\sqrt{3}=2$
this is not correct

maroonblazer
Posts: 51
Joined: Thu Aug 12, 2010 11:16 am
Contact:

### Re: Help with Solving an Equation Containing Radicals...

Hi,
I've got the following problem:
$y\sqrt{3} + 1 = 3 - y$

I add y to both sides to rid y on the right. Then I subtract 1 from both sides to get this:

$2y\sqrt{3}=2$
this is not correct
Ah, ok, thanks. Instead, I think I should have done:
$y + y\sqrt{3} = 2$
$y(1 + \sqrt{3}) = 2$

Multiply the numerator and denominator by the conjugate, $1-\sqrt{3}$, to get:
$\frac{2-\sqrt{3}}{-2}$

which simplifies to:
$-1+\sqrt{3}$

Thanks again!
mb

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
Instead, I think I should have done:
$y + y\sqrt{3} = 2$
$y(1 + \sqrt{3}) = 2$

Multiply the numerator and denominator by the conjugate, $1-\sqrt{3}$...
What "numerator and denominator"? Or do you mean that you first divided through to get $y$ by itself? If so, then:

. . . . .$y\left(1\, +\, \sqrt{3}\right)\, =\, 2$

. . . . .$y\, =\, \frac{2}{1\, +\, \sqrt{3}}$

Then you did the multiplication by $1\, -\, sqrt{3}$ to get:

. . . . .$y\, =\, \left(\frac{2}{1\, +\, \sqrt{3}}\right)\left(\frac{1\, -\, \sqrt{3}}{1\, -\, \sqrt{3}}\right)$
$\frac{2-\sqrt{3}}{-2}$
How did the $2$ multiply only on the $1$ and not also the $\sqrt{3}\,?$

maroonblazer
Posts: 51
Joined: Thu Aug 12, 2010 11:16 am
Contact:

### Re:

Instead, I think I should have done:
$y + y\sqrt{3} = 2$
$y(1 + \sqrt{3}) = 2$

Multiply the numerator and denominator by the conjugate, $1-\sqrt{3}$...
What "numerator and denominator"? Or do you mean that you first divided through to get $y$ by itself? If so, then:

. . . . .$y\left(1\, +\, \sqrt{3}\right)\, =\, 2$

. . . . .$y\, =\, \frac{2}{1\, +\, \sqrt{3}}$

Then you did the multiplication by $1\, -\, sqrt{3}$ to get:

. . . . .$y\, =\, \left(\frac{2}{1\, +\, \sqrt{3}}\right)\left(\frac{1\, -\, \sqrt{3}}{1\, -\, \sqrt{3}}\right)$
$\frac{2-\sqrt{3}}{-2}$
How did the $2$ multiply only on the $1$ and not also the $\sqrt{3}\,?$
Sorry, yes, I skipped a step above. And you're right - I should've multiplied on the $\sqrt{3}$ too.

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