Question

Where am I going wrong here?

6/x - 5 = x

6 - 5x = x^2

0 = x2 + 5x -6

-5/2 +- sqrt((5/2)^2 + 6)

-5/2 +- sqrt(49)

-5/2 +- 7

-5/2 +- 14/2

wrong... answers should be x: 1, x: -6

Answer 1

Is the problem 6/(x-5) or (6/x)-5?

Answer 2

You need to move the -5 over first, and then make it into a quadratic and factor

Answer 3

6/x = x+5
6= x^2+5
x^2 +5 -6 = 0
(x-1)(x+6)=0

Answer 4

yep ^

Answer 5

How did you find -1 and 6 though...?

Answer 6

By factoring the quadratic

Answer 7

(x-1)(x+6)=0
x = 1, -6

Answer 8

I don't see how you get this though: (x-1)(x+6)=0

Answer 9

I'm sure deybow is typing up a long factoring explanation for you :)

Answer 10

See that's what I tried to do with the (overly complicated?) formula, but I keep getting the wrong answer.

Answer 11

If you want you can use the quadratic formula, but there are faster factoring methods

Answer 12

This is the formula I tried to use:

x^2 + px +q = 0

x = -p/2 +- sqrt((p/2)^2 -q)

Answer 13

\[(-b \pm \sqrt(b^2 - 4ac))/2a\]

Answer 14

Basically at this point you have x^2+5x-6 and you have to factor it. When you do factor it, you get (x+6)(x-1).

Since it's quadratic, or since it has an x^2 in it, you know that there will be two separate parts in the answer that will each have an x in them, so you can start out by writing (x+-_)(x+-_) because the two x's would multiply together to make the x^2. Hopefully this makes sense so far. And the reason why it's + or - is because you have to figure out later which side has to be added or subtracted.

Now the _'s in each set of parenthesis have to be numbers that will multiply to get -6 and add together to get 5. These numbers happen to be 6 and -1. So, you would put (x+6)(x-1). Now remember, this is all equal to zero, so the solutions will then be x=-6 and 1 because either of those values would make the equation come out to zero.

Hopefully that makes sense :P

Answer 15

Told you he was typing up a nice long explanation for you :)

Answer 16

haha it's hard to explain actually :\

Answer 17

it is, but it's a very ambitious answer! very much appreciated

Answer 18

ha thanks.

Answer 19

I'll try your formula instead: (−b±(√b2−4ac))/2a

I still *think* the one I used should work, but, idk.

Answer 20

thanks, both of you!

Answer 21

Yeah the quadratic formula is the longer method, and also the way you have to calculate the answer for any quadractics that can't be factored the easy way.

Answer 22

But if you have any questions about the shortcut method I can try to explain more :P

Answer 23

I need to learn the formulas first, I'll get to shortcuts when I've got this down. ;) thanks though