# [SOLVED] Quadratic Equation

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• May 2nd 2008, 05:35 AM
looi76
[SOLVED] Quadratic Equation
Question:
Find the value of $x$ when $y = 0$
$y = \frac{2x - 1}{x + 3}$

Attempt:

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

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

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

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

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

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

I think the quadratic equation that i got is wrong (Headbang)... Where did I go wrong?
• May 2nd 2008, 05:42 AM
Moo
Hello,

Quote:

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

Attempt:

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

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

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

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

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

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

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...

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

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

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

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

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

:)

But I don't see why there is a quadratic equation here... ^^
• May 2nd 2008, 05:42 AM
mr fantastic
Quote:

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

Attempt:

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

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

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

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

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

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

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, 05:53 AM
zp3929
Quote:

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

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

$
{2x - 1}= 0
$

$
{2x} = 1
$

$
{x} = 1/2
$

Zac
• May 2nd 2008, 05:55 AM
Moo
Quote:

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

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

$
{-x + 4} = 0
$

$
{-x} = -4
$

$
{x} = 4
$

Zac

What you did was for y=1 (Wink)
• May 2nd 2008, 06:00 AM
Moo
Quote:

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

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

$
{-3x -2} = 0
$

$
{-3x} = 2
$

$
{x} = -2/3
$

Zac

Now, this is for y=-1 (Giggle)
• May 2nd 2008, 06:03 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, 06:11 AM
mr fantastic
Quote:

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

$
{2x - 1}= 0
$

$
{2x} = 1
$

$
{x} = 1/2
$

Zac

(Clapping)
• May 2nd 2008, 06:12 AM
looi76
$y = \frac{2x - 1}{x + 3}$

$y = 0$

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

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

The answer I have is $x = \frac{1}{2}$ ?!(Crying)
• May 2nd 2008, 06:14 AM
looi76
Quote:

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

$
{2x - 1}= 0
$

$
{2x} = 1
$

$
{x} = 1/2
$

Zac

Quote:

Originally Posted by mr fantastic
(Clapping)

Why do we get rid of $x + 3$ ?
• May 2nd 2008, 06:15 AM
zp3929
Quote:

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

$y = 0$

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

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

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

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

we get rid of the $x = 3$ because $x = 3$/0 = 0 which leaves us with $2x - 1 = 0$
• May 2nd 2008, 06:18 AM
looi76
Quote:

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

Thanks zp3929! I couldn't understand the reason for you removing $x + 3$ from the equation?
• May 2nd 2008, 01:06 PM
Moo
Quote:

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

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