# Solve and functions

Printable View

• Dec 14th 2006, 04:28 AM
Naif
Solve and functions
(1/x - 2) = (3/x + 2) - (6x/x² - 4)

and

f(x) = √5x² + 3x - 1

and

xª + yª = zª

can anyone explain or run me through any of these :S i been up all night trying to do them
• Dec 14th 2006, 04:53 AM
topsquark
Quote:

Originally Posted by Naif
(1/x - 2) = (3/x + 2) - (6x/x² - 4)

and

f(x) = √5x² + 3x - 1

and

xª + yª = zª

can anyone explain or run me through any of these :S i been up all night trying to do them

CaptainBlack showed in your other post that the first one contains a contradiction, so something is wrong with it.

The second one is a function, not a question. What are you asking about it?

The last one looks similar to Fermat's Last Theorem, but other than that you again have asked no question about it.

-Dan
• Dec 14th 2006, 04:59 AM
Naif
first one, must be wrong then i saw the contradiction aswell but i thought id ask ... second one i have to solve for x ... and third one gotta make xª + yª equal to
zª and a has to be larger then 1
• Dec 14th 2006, 05:35 AM
CaptainBlack
Quote:

Originally Posted by Naif
first one, must be wrong then i saw the contradiction aswell but i thought id ask ... second one i have to solve for x ... and third one gotta make x&#170; + y&#170; equal to
z&#170; and a has to be larger then 1

x^a+b^a=z^a,

let a=2, x=1, b=1, then the left hand side is:

1^2+1^2=2,

so let z=sqrt(2) which gives a solution.

But that is unlikely to be the question you were asked.

You won't get much help with questions expressed like these, as they are
there is more work in turning then into questions that in solving them. Not
to mention we are wasting our time, as our interpretations are unlikely
to be what you were actually asked.

In future try to give the question exactly as asked (which includes all
those inessential words like: show, find, solve, simplify, .. they do mean
something)

Also don't make multiple posts of the same questions, and don't delete
a question once its been asked, you are confusing the thread structure,
and wasting the helpers time.

RonL
• Dec 14th 2006, 05:34 PM
Naif
f(x) = √5x² + 3x - 1

I have to solve for "x"
Still am lost on how to do it tho its been a while since ive done it
• Dec 14th 2006, 10:33 PM
CaptainBlack
Quote:

Originally Posted by Naif
f(x) = √5x&#178; + 3x - 1

I have to solve for "x"
Still am lost on how to do it tho its been a while since ive done it

That still does not make any sense, what you have asked is for us to find the
point of intersection of an arbitary curve y=f(x), and the parabola y=√5x&#178; + 3x - 1
which cannot be done without knowing something about f(x).

What your question probably asks is something like: solve f(x)=0, where
f(x) = √5x&#178; + 3x - 1.

For what it's worth:

solve:

√5x&#178; + 3x - 1=0.

This is a quadratic in x so we use the quadratic formula:

x = [-3 +/- sqrt(9 + 4√5)]/(2√5) ~= 0.305626 and -1.647267

RonL