Melody

This question does not make any sense to me.  

this site might be useful too :)

If you need to solve by hand then use the quadratic formula

Use the calc on the home page:)

I am not a all sure what you want.

Can't you gather at lease one of the examples and we can talk about it.    

2*x^2-(9*x)+4

$$\\2x^2-9x+4\\\\$$

You need to find 2 numbers that multiply to 2*4=8  and add togeter to give -9

That is easy  -8 and -1

Now split -9x into  -8x - 1x

like this

$$\\2x^2-8x-1x+4\\\\$$

now group them in pairs like this

$$\\2x(x-4)-1(x-4)\\\\$$

Now you have 2x lots of (x-4) take away 1 lot of (x-4)

that is 

$$\\(2x-1)(x-4)\\\\$$

Hi Alan,

Yes I did not think that L'Hopital's rule helped either.

I was just pointing out that the question requested it and it was not used.  That is all.   :)

When y=2

$$4x^2=1+4/9$$

$${\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}} = {\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{4}}}{{\mathtt{9}}}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{13}}}}}{{\mathtt{6}}}}\\
{\mathtt{x}} = {\frac{{\sqrt{{\mathtt{13}}}}}{{\mathtt{6}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.600\: \!925\: \!212\: \!577\: \!331\: \!5}}\\
{\mathtt{x}} = {\mathtt{0.600\: \!925\: \!212\: \!577\: \!331\: \!5}}\\
\end{array} \right\}$$

width at the top =  $${\frac{{\sqrt{{\mathtt{13}}}}}{{\mathtt{3}}}}$$    metres

It appears to me that we all got different answers.  

I for one did not really understand the question.  

Now Chris's is the same as mine (he did a little correction) but Heueka's is still different.

Maybe Heurka  interpreted it differently from me and Chris?

Well I can demonstrate with a pattern

4*3    =12

4*2     =8

4*1     =4

4*0     =0

4*-1    =-4

4*-2    =-8

Can you see that the pattern shows that a pos * a neg = a neg  ?

Now i will swap it around

-2*4   =-8

-2*3  =-6

-2*2  =-4

-2*1  =-2

-2*0  =0

-2*-1  =+2

-2*-2  =+4

-2*-3   =+6

Can you see how the pattern shows that a neg * a neg = a pos  ?

there are other ways to show this as  well.  One of the other answerers may show you another way. :)

$${\frac{\left({{\mathtt{10}}}^{{\mathtt{13}}}\right)}{\left({{\mathtt{10}}}^{{\mathtt{10}}}\right)}} = {\mathtt{1\,000}}$$

1000 years