Melody

Fri/Sat 5/7/14

♬                                        ♬ ♬ MELODY ♬ ♬

Sum and product of quadratic roots

Here is another approach.  I suspect that this is how it was meant to be done.

Given the quadratic equation 

$$ax^2+bx+c=0\qquad \mbox{the sum of the roots is } \frac{-b}{a} \mbox{ and the product of the roots is } \frac{c}{a}$$

so

$$2x^2-5x-7=0\qquad r+s= \frac{5}{2}\:\: \mbox{ and }\:\: rs=\frac{-7}{2}$$

Now

$$\dfrac{r}{s}+\dfrac{s}{r}\\\\
=\dfrac{r^2+s^2}{rs}\\\\
=\dfrac{r^2+s^2+2rs-2rs}{rs}\\\\
=\dfrac{(r+s)^2-2rs}{rs}\\\\
=\dfrac{(\frac{5}{2})^2-2\times \frac{-7}{2}}{\frac{-7}{2}}\\\\
=\dfrac{\frac{25}{4}+ \frac{14}{2}}{\frac{-7}{2}}\\\\
=\dfrac{\frac{25}{4}+ \frac{28}{4}}{\frac{-7}{2}}\\\\
=\frac{53}{4}\times\frac{-2}{7}\\\\
=\frac{53}{2}\times\frac{-1}{7}\\\\
=\frac{-53}{14}\\\\
=\:-3\frac{11}{14}$$

Thank you for that wonderful accolade David.

I do not think for one moment that I deserve it. Nor do I think I can live up to it.

But thank you so much. :-)

Thanks Alan,

Maybe Zegroes or another young US person could tell us about how US people celebrate July 4.

Yes Rosala.  It is fun.  It is nice to have such great friends too. 

No it looks ok this time.

See, all my nagging is worth while.

Thanks Rosala,

What did I say, I have forgotten.

Did you see you got one right that Zegroes go wrong.  Chris got it wrong too because he said you were both correct!    

There wasn't a balanced number of brackets so I added one at the end.

$$((2^{1/2})*(2-3^{1/2})^{1/2}+3)^{-1/2}-(1-(2^{1/2})*((2-3^{1/2})^{1/2})^{1/2})\\\\
=((2^{1/2})*(2-3^{1/2})^{1/2}+3)^{-1/2}-(1-(2^{1/2})(2-3^{1/2})^{1/4})\\\\
=\frac{1}{((2^{1/2})*(2-3^{1/2})^{1/2}+3)^{1/2}}-(1-(2^{1/2})(2-3^{1/2})^{1/4})\\\\
=\dfrac{1}{\sqrt{\sqrt2 *\sqrt{(2-\sqrt3)}+3}}-\left(1-\sqrt2 \:\sqrt[4]{(2-\sqrt3}\:\right)\\\\$$

The site calcanswered what you wrote.  I answered something closer to what you may have meant.

I doubt that it was a serious question anyway so it really doesn't matter.   

$${\left({\frac{\left({\frac{{{\mathtt{2}}}^{{\mathtt{1}}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{\left({\mathtt{2}}{\mathtt{\,-\,}}{\frac{{{\mathtt{3}}}^{{\mathtt{1}}}}{{\mathtt{2}}}}\right)}^{{\mathtt{1}}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}^{{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}}{\mathtt{\,-\,}}\left({\mathtt{1}}{\mathtt{\,-\,}}{\frac{\left({\frac{{{\mathtt{2}}}^{{\mathtt{1}}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{\left({\frac{{\left({\mathtt{2}}{\mathtt{\,-\,}}{\frac{{{\mathtt{3}}}^{{\mathtt{1}}}}{{\mathtt{2}}}}\right)}^{{\mathtt{1}}}}{{\mathtt{2}}}}\right)}^{{\mathtt{1}}}}{{\mathtt{2}}}}\right) = -{\mathtt{0.320\: \!299\: \!803\: \!774\: \!770\: \!9}}$$

@@ End of Day Wrap - Wed/Thurs 3/6/14   Sydney, Australia   Time 11:30am       ♬              

Hi all,                                                                             ♬

Lots of great answers today from Rosala, Zegroes, CPhill, Jboy314, admin, Alan, SevenUP, Heureka and dragonSlayer554.  Thanks everyone.      ♬                                                                           ♬ 

Yesterday Zegroes earned his star.                     CONGRATULATIONS ZEGROES     

Rose98 and DragonSlayer554 are both new.  Welcome to web2.0calc forum.  We hope that you learn lots and enjoy yourselves here.                                                      ♬

I worked out how we can duck between the different language forums without having to sign in each time.    

I hae a German post in my watchlist and if i use this to gain access i can go to the german site without any hassles and then I can go back again via an English post in my watch list.  It is really cool!

This is the German post that i used.  I assume that it will work from there as well but I am not sure.

Why don't you google this and come back and tell us the answer ?