Melody

Hi Xxmathsguy,

Welcome to our Web2.0calc forum.   I can see you have been busy here.  I hope you are enjoying yourself.   

You do not need to appologise to the guest.  Or to anyone for that matter.

Askers should always check the answers.

If they are too lazy or have so little knowledge that they can't then they are just copying and they should get it wrong.

If they are unsure then they should ask for further clarification.

I would give you a hint if I had produced anything that made sense.  

I'll start by stating that I do not have a clue.

Now the discussion:

when x=0     k can be anything

But, for any other given x,  k cannot equal 2/x     

But I don't know what the answer is.     I tried graphing it but that just confused me more.  

Maybe it is    \((-\infty, \frac{2}{x}) \cup (\frac{2}{x},\infty)\)    where   \(x\ne 0\)           But I am guessing.  

You made a boo boo.

It is just a careless error but you should have noticed that your answer does not make sense.

Why don't you put an ounce of effort into presenting your question properly?

I answered this or something very similar a few days ago.  Go see if you can find it.

Added:

I just googled you exact question and found an anwer immediately.

x = 1 - sqrt(2)] and x = 2 + sqrt(7).

\(f(x)=(x-(1-\sqrt2))(x-(1+\sqrt2))(x-(2-\sqrt7))(x-(2+\sqrt7))\)

That should do it.  You can expand.  Try graphing it using desmos to check it is right first ;)

What a load of rubbish.
If they were given to you as a fun challenge and you are trying to get them done here then you are cheating on your freinds challenge.

More likely you are cheating on school work.

Either way, your aim with this post is to CHEAT!

Your cheat questions are blocked, not that that anyone would be likely to bother with them anyway.

\((1 + x)^{10}*(1 - x)^{10}\\ =[(1 + x)(1 - x)]^{10}\\ =[1-x^2]^{10} \)

This will have 11 terms

\(1-10x^2+ ..... -10x^{18}+x^{20}\)

\(3x+7\equiv 2+x+5\;\;mod 16)\\ 2x\equiv 0 (mod 16)\\ x\equiv 8N\\ ~\\ 2x+11+x+3\;\; (mod16) \\ \equiv 0+x+14\;\; (mod16) \\ \equiv 8N-2 \;\; (mod16) \\ \)

So I think  \(2x+11+x+3(mod 16) \equiv 6\;(mod 16) \quad or \quad 14\;(mod 16) \)

I am not 100% certain that this is correct.     Please give feedback.

LaTex

3x+7\equiv  2+x+5\;\;mod 16)\\
2x\equiv  0 (mod 16)\\
x\equiv  8N\\
~\\
2x+11+x+3\;\; (mod16) \\
\equiv  0+x+14\;\; (mod16) \\
\equiv  8N-2 \;\; (mod16) \\