[SOLVED] Quadratic Equation

looi76 · #1 $Old$ May 2nd, 2008, 05:35 AM

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

... Where did I go wrong?

Moo · #2 $Old$ May 2nd, 2008, 05:42 AM

Hello,

Quote:

Originally Posted by looi76 $View Post$

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

... 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... ^^

mr fantastic · #3 $Old$ May 2nd, 2008, 05:42 AM

Quote:

Originally Posted by looi76 $View Post$

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

... 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?)

zp3929 · #4 $Old$ May 2nd, 2008, 05:53 AM

Quote:

Originally Posted by looi76 $View Post$

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

Moo · #5 $Old$ May 2nd, 2008, 05:55 AM

Quote:

Originally Posted by zp3929 $View Post$

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

Moo · #6 $Old$ May 2nd, 2008, 06:00 AM

Quote:

Originally Posted by zp3929 $View Post$

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

zp3929 · #7 $Old$ May 2nd, 2008, 06:03 AM

Quote:

Originally Posted by Moo $View Post$

Now, this is for y=-1

i suck at this today
ok 1 last try

mr fantastic · #8 $Old$ May 2nd, 2008, 06:11 AM

Quote:

Originally Posted by zp3929 $View Post$

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

looi76 · #9 $Old$ May 2nd, 2008, 06:12 AM

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

$y = 0$

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

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

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

looi76 · #10 $Old$ May 2nd, 2008, 06:14 AM

Quote:

Originally Posted by zp3929 $View Post$

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 $View Post$

Why do we get rid of $x + 3$ ?

zp3929 · #11 $Old$ May 2nd, 2008, 06:15 AM

Quote:

Originally Posted by looi76 $View Post$

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

$y = 0$

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

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

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

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