• May 2nd 2008, 04:35 AM
looi76
Question:
Find the value of $\displaystyle x$ when $\displaystyle y = 0$
$\displaystyle y = \frac{2x - 1}{x + 3}$

Attempt:

$\displaystyle (2x - 1)(x + 3)^{-1} = 0$

$\displaystyle (2x - 1)(x^{-1} + \frac{1}{3}) = 0$

$\displaystyle 2 + \frac{2}{3}x - x^{-1} - \frac{1}{3} = 0$

$\displaystyle \frac{2}{3}x - x^{-1} + \frac{5}{3} = 0$

$\displaystyle \frac{4}{9}x^2 - 1 + \frac{5}{3} = 0$

$\displaystyle \frac{4}{9}x^2 + \frac{5}{3} - 1 = 0$

I think the quadratic equation that i got is wrong (Headbang)... Where did I go wrong?
• May 2nd 2008, 04:42 AM
Moo
Hello,

Quote:

Originally Posted by looi76
Question:
Find the value of $\displaystyle x$ when $\displaystyle y = 0$
$\displaystyle y = \frac{2x - 1}{x + 3}$

Attempt:

$\displaystyle (2x - 1)(x + 3)^{-1} = 0$

$\displaystyle {\color{red}(2x - 1)(x^{-1} + \frac{1}{3}) = 0}$

$\displaystyle 2 + \frac{2}{3}x - x^{-1} - \frac{1}{3} = 0$

$\displaystyle \frac{2}{3}x - x^{-1} + \frac{5}{3} = 0$

$\displaystyle \frac{4}{9}x^2 - 1 + \frac{5}{3} = 0$

$\displaystyle \frac{4}{9}x^2 + \frac{5}{3} - 1 = 0$

I think the quadratic equation that i got is wrong (Headbang)... Where did I go wrong?

It's the step in red which is incorrect...

$\displaystyle (a+b)^n \not = a^n+b^n$

What you have to do is to say that x+3 has to be different of 0.
Then solve :

$\displaystyle 0=\frac{2x-1}{x+3}$

Multiplying both sides by x+3 (which is possible because x+3 is not = 0) :

$\displaystyle \boxed{0=2x-1}$

:)

But I don't see why there is a quadratic equation here... ^^
• May 2nd 2008, 04:42 AM
mr fantastic
Quote:

Originally Posted by looi76
Question:
Find the value of $\displaystyle x$ when $\displaystyle y = 0$
$\displaystyle y = \frac{2x - 1}{x + 3}$

Attempt:

$\displaystyle (2x - 1)(x + 3)^{-1} = 0$

$\displaystyle (2x - 1)(x^{-1} + \frac{1}{3}) = 0$

$\displaystyle 2 + \frac{2}{3}x - x^{-1} - \frac{1}{3} = 0$

$\displaystyle \frac{2}{3}x - x^{-1} + \frac{5}{3} = 0$

$\displaystyle \frac{4}{9}x^2 - 1 + \frac{5}{3} = 0$

$\displaystyle \frac{4}{9}x^2 + \frac{5}{3} - 1 = 0$

I think the quadratic equation that i got is wrong (Headbang)... Where did I go wrong?

You've made very heavy weather of this. Simply solve 2x - 1 - 0 (why?).

And it's good form to check that the solution is not a solution to x + 3 = 0 (why?)
• May 2nd 2008, 04:53 AM
zp3929
Quote:

Originally Posted by looi76
Question:
Find the value of $\displaystyle x$ when $\displaystyle y = 0$
$\displaystyle y = \frac{2x - 1}{x + 3}$

this is what i did (dont know if it is right or not though)
$\displaystyle y = \frac{2x - 1}{x + 3}$

$\displaystyle {2x - 1}= 0$

$\displaystyle {2x} = 1$

$\displaystyle {x} = 1/2$

Zac
• May 2nd 2008, 04:55 AM
Moo
Quote:

Originally Posted by zp3929
this is what i did (dont know if it is right or not though)
$\displaystyle y = \frac{2x - 1}{x + 3}$

$\displaystyle {x + 3} = {2x - 1}$

$\displaystyle {-x + 4} = 0$

$\displaystyle {-x} = -4$

$\displaystyle {x} = 4$

Zac

What you did was for y=1 (Wink)
• May 2nd 2008, 05:00 AM
Moo
Quote:

Originally Posted by zp3929
this is what i did (dont know if it is right or not though)
$\displaystyle y = \frac{2x - 1}{x + 3}$

$\displaystyle {-x - 3} = {2x - 1}$

$\displaystyle {-3x -2} = 0$

$\displaystyle {-3x} = 2$

$\displaystyle {x} = -2/3$

Zac

Now, this is for y=-1 (Giggle)
• May 2nd 2008, 05:03 AM
zp3929
Quote:

Originally Posted by Moo
Now, this is for y=-1 (Giggle)

i suck at this today
ok 1 last try
• May 2nd 2008, 05:11 AM
mr fantastic
Quote:

Originally Posted by zp3929
this is what i did (dont know if it is right or not though)
$\displaystyle y = \frac{2x - 1}{x + 3}$

$\displaystyle {2x - 1}= 0$

$\displaystyle {2x} = 1$

$\displaystyle {x} = 1/2$

Zac

(Clapping)
• May 2nd 2008, 05:12 AM
looi76
$\displaystyle y = \frac{2x - 1}{x + 3}$

$\displaystyle y = 0$

$\displaystyle \frac{2x - 1}{x + 3} = 0$

$\displaystyle 2x - 1 = x + 3$
$\displaystyle 2x - 1 - x - 3 = 0$
$\displaystyle x - 4 = 0$
$\displaystyle x = 4$

The answer I have is $\displaystyle x = \frac{1}{2}$ ?!(Crying)
• May 2nd 2008, 05:14 AM
looi76
Quote:

Originally Posted by zp3929
this is what i did (dont know if it is right or not though)
$\displaystyle y = \frac{2x - 1}{x + 3}$

$\displaystyle {2x - 1}= 0$

$\displaystyle {2x} = 1$

$\displaystyle {x} = 1/2$

Zac

Quote:

Originally Posted by mr fantastic
(Clapping)

Why do we get rid of $\displaystyle x + 3$ ?
• May 2nd 2008, 05:15 AM
zp3929
Quote:

Originally Posted by looi76
$\displaystyle y = \frac{2x - 1}{x + 3}$

$\displaystyle y = 0$

$\displaystyle \frac{2x - 1}{x + 3} = 0$

$\displaystyle 2x - 1 = x + 3$
$\displaystyle 2x - 1 - x - 3 = 0$
$\displaystyle x - 4 = 0$
$\displaystyle x = 4$

The answer I have is $\displaystyle x = \frac{1}{2}$ ?!(Crying)

yer it is $\displaystyle x = \frac{1}{2}$
i edited my post (several times(Itwasntme))

we get rid of the $\displaystyle x = 3$ because $\displaystyle x = 3$/0 = 0 which leaves us with $\displaystyle 2x - 1 = 0$
• May 2nd 2008, 05:18 AM
looi76
Quote:

Originally Posted by zp3929
yer it is $\displaystyle x = \frac{1}{2}$
i edited my post (several times(Itwasntme))

Thanks zp3929! I couldn't understand the reason for you removing $\displaystyle x + 3$ from the equation?
• May 2nd 2008, 12:06 PM
Moo
Quote:

Originally Posted by looi76
Thanks zp3929! I couldn't understand the reason for you removing $\displaystyle x + 3$ from the equation?

Told you, it's because you multiply each side by x+3 ^^