Quadratic.

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July 31st 2009, 05:17 PM
el123

Quadratic.

ok so i have this problem.

$\frac{x+4}{x-2} - \frac{x-3}{x-4}$

So i know i should multiply both sides by the denominator then factorize then subtract , then use quadratic formula right?

But i keep getting mixed up , or maybe i am missing somthing. Can somebody please help me.
July 31st 2009, 05:48 PM
Kasper

Quote:

Originally Posted by el123

ok so i have this problem.

$\frac{x+4}{x-2} - \frac{x-3}{x-4}$

So i know i should multiply both sides by the denominator then factorize then subtract , then use quadratic formula right?

But i keep getting mixed up , or maybe i am missing somthing. Can somebody please help me.

$\frac{x+4}{x-2} - \frac{x-3}{x-4}=0$ ?

To use the quadratic function, one side must be equal to 0. Did your question not include an equality? You can't find the value of x without having the function equal something (and if you did have any value instead of 0, you could bring it to the x side to get 0); Even if you simplify that down to the form $ax^2 + bx +c$ , you don't know what that function is equal to, x could have an infinite number of possibilities. Otherwise I'd say your instructor wants you to assume it is equal to 0 and do as you are planning to.
July 31st 2009, 05:53 PM
el123

yeah my bad i forgot , yes it is equal to zero.
July 31st 2009, 05:55 PM
Chris L T521

Quote:

Originally Posted by el123

ok so i have this problem.

$\frac{x+4}{x-2} - \frac{x-3}{x-4}$

So i know i should multiply both sides by the denominator then factorize then subtract , then use quadratic formula right?

But i keep getting mixed up , or maybe i am missing somthing. Can somebody please help me.

See this thread: http://www.mathhelpforum.com/math-he...-question.html

Turns out to be a very similar problem.
August 1st 2009, 06:03 AM
stapel

Quote:

Originally Posted by el123

...i keep getting mixed up , or maybe i am missing somthing. Can somebody please help me.

Certainly! Please reply showing what you've tried, and we'll be glad to help you find where you're getting mixed up or missing something. Thank you! (Wink)
August 1st 2009, 06:35 AM
Stroodle

In case you haven't worked it out yet:

$\frac{x+4}{x-2}-\frac{x-3}{x-4}=0$

$(x+4)(x-4)-(x-3)(x-2)=0$

$(x^2-16)-(x^2-5x+6)=0$

$5x-22=0$

$x=\frac{22}{5}$