Maths, help pleaseeeee

Les_Rebels123

THIS ONE IS SOLVED, thanks to pathway33, but there is another, im stuck on, same context,just twisted. Please scroll down to end and if you can help/answer, it'd be great Thanks In Advance


Solve the equation: x^2-3x-28=0
right ive done that x=+7 or x=-4
BUT

Hence solve the equation: (2y-1)^2 - 3(2y-1)-28=0

help anyone.thanks in advance

^means to the power of :P

pathway33

x = 2y - 1

so 2y - 1 = 7 or 2y - 1 = -4

2y = 8 or 2y = -3

y = 4 or -1.5

Les_Rebels123

ok thanks

but another one,, help would be appreciated once more

x^2-9x+20=0
x=+5 or x=+4

hence solve fully the equation

(y+[4/y])^2 -9(y+[4/y]) +20=0

Are you sure you've taken it down correctly? Specifically, in the first equation you've a -20, while in the second you've a +20. Is this correct?

And, I'm not going to give you the full solution, where exactly are you having difficulty?

pathway33

Les_Rebels123 wrote: »

ok thanks

but another one,, help would be appreciated once more

x^2-9x-20=0
x=+5 or x=+4

hence solve fully the equation

(y+[4/y])^2 -9(y+[4/y]) +20=0

So your first equation should read x^2 - 9x + 20 = 0.....cos -20 don't make sense

So, x = y + [4/y]

Therefore y + [4/y] = 4 or y + [4/y] = 5

Taking y + [4/y] = 4, we multiply the whole equation by the LCM which is y.

This gives us y^2 + 4 = 4y

y^2 - 4y + 4 = 0

(y - 2)(y - 2) = 0

y = 2

Taking y + [4/y] = 5, we multiply the whole equation by the LCM which is y.

This gives us y^2 + 4 = 5y

y^2 - 5y + 4 = 0

(y - 1)(y - 4) = 0

y = 1 or y = 4

ANSWER y = 1, 2 or 4

Les_Rebels123

gettn the values for y...

x=y+4/y
where to from here?

Les_Rebels123

Ohhh.... I get it now thanks so much...hopefully I wont have to ask any more on these equations. When its all laid out clearly,I can see where things come from, and where to go next. when someone just writes the answer when explaining its just not easy to Make out...
And, guys, im not being lazy by asking you guys to do my h/w, i just needed it explained better :P as i had tried several diff ways,writing in pencil,getting the wrong answer,erasing out,trying again,but to no avail

thanks once again
L.Rebel

The solutions are, as pathway has shown, y = 1, 2 or 4.

Here's where it comes from:

You know that for [latex]\displaystyle\mbox{x}^2 - 9x + 20 = 0[/latex], x = 4 or x = 5.

Now, compare [latex]\displaystyle\mbox{x}^2 - 9x + 20 = 0[/latex] to [latex]\displaystyle\mbox{\left(y + \frac{4}{y}\right)^2} + 9\mbox{\left(y + \frac{4}{y}\right) + 20 = 0[/latex]. If you compare them, you'll see that x and [latex]\displaystyle\mbox{\left(y + \frac{4}{y}\right)}[/latex] are equal, i.e. they're the same equations, [latex]\displaystyle\mbox{(blah)}^2 + 9\mbox{(blah)} + 20 = 0[/latex], only the "blah" is different on both occasions.

Now, since you know that x = [latex]\displaystyle\mbox{\left(y + \frac{4}{y}\right)}[/latex], you can let the latter equal to 4 or to 5, since you know that x = 4 or 5. You then let this [latex]\displaystyle\mbox{\left(y + \frac{4}{y}\right)}[/latex] equal to 4 and 5, you'll get two equations, and you solve both, giving you your solutions of 1, 2 and 4.

(P.S. I know pathway has given you full solutions, but they're easier to understand when they're written in LaTeX, I think).

Les_Rebels123

Both Ways are understandable,except for the way i was shown off site :P
you guys have realllllly reallllllly helped me.

Thanks to both of you, and I have thanked both of you tooo

-JammieDodger- Is LaTeX a free program, or does it have to be paid for. Im assuming that one would pay for such an application/program. :P

Les_Rebels123

Les_Rebels123 wrote: »

-JammieDodger- Is LaTeX a free program, or does it have to be paid for. Im assuming that one would pay for such an application/program. :P

It's free, and it's built into Boards. All you have to do is wrap a certain code with [noparse][latex][/latex][/noparse] tags. If you quote my above posts, you'll see some of the code. Also, have a look in the Mathematics forum, there's a few guides over there.

stesh

Les_Rebels123 wrote: »

Is LaTeX a free program, or does it have to be paid for. Im assuming that one would pay for such an application/program. :P

lol. LaTeX is quite famous for being free...

http://miktex.org/ for Windows, or http://www.tug.org/mactex/ for OS X are your best bets. If you run Linux etc. you can get it through a package manager.

Les_Rebels123

so,*testing* [latex]x^2[/latex]

thanks again, ill have to favourite/bookmark/write it down, handy to know these things,as so it wont make the equation look complicated cheeeeerrrrsss

Les_Rebels123

stesh wrote: »

lol. LaTeX is quite famous for being free...

http://miktex.org/ for Windows, or http://www.tug.org/mactex/ for OS X are your best bets. If you run Linux etc. you can get it through a package manager.

Thankkkkkkssss Downloading now

Les_Rebels123

So, Should I post the Thread In Leaving Cert, Or on the Mathematics one?? or does it make a difference? either or?

Les_Rebels123

Les_Rebels123 wrote: »

So, Should I post the Thread In Leaving Cert, Or on the Mathematics one?? or does it make a difference? either or?

If it's a thread about homework problems, here's your best bet I'd say. The Maths forum can get pretty quiet.