Complex Surds

**j0k3r** · September 22nd 2008, 10:49 AM

Hi everyone. Just found the forum. Revising over surds etc and have hit a snag. Can't figure out how to answer this question:

You are given that:

$a = \frac{3}{2}$ , $b = \frac{9-\sqrt{17}}{4}$ and $c = \frac{9+\sqrt{17}}{4}$

Show that $a + b + c = abc$ .

Thanks in advance for any help... I'm tearing my hair out over this one.

Mo.

**j0k3r** · September 22nd 2008, 10:51 AM

Therefore:

$\frac{3}{2} + \frac{9-\sqrt{17}}{4} + \frac{9+\sqrt{17}}{4} = \frac{3}{2}*\frac{9-\sqrt{17}}{4}*\frac{9+\sqrt{17}}{4}$ .

Mo.

**j0k3r** · September 22nd 2008, 10:56 AM

$\frac{3}{2} + \frac{9-\sqrt{17}}{4} + \frac{9+\sqrt{17}}{4} = \frac{6*18-17}{4}$

$\frac{3}{2} + \frac{9-\sqrt{17}}{4} + \frac{9+\sqrt{17}}{4} = 91$

right? or is it:

$\frac{3}{2} + \frac{9-\sqrt{17}}{4} + \frac{9+\sqrt{17}}{4} = \frac{6*18*-17}{4}$

$\frac{3}{2} + \frac{9-\sqrt{17}}{4} + \frac{9+\sqrt{17}}{4} = -1836$

aghhhh >_<

**Moo** · September 22nd 2008, 11:01 AM

Originally Posted by j0k3r

$\frac{3}{2} + \frac{9-\sqrt{17}}{4} + \frac{9+\sqrt{17}}{4} = \frac{6*18-17}{4}$

Hmmm no !

$\frac 32+\frac{9-\sqrt{17}}{4}+\frac{9+\sqrt{17}}{4}=\frac 64+\frac{9-\sqrt{17}}{4}+\frac{9+\sqrt{17}}{4}$

Getting everything on a unique fraction :

$=\frac{6+(9-\sqrt{17})+(9+\sqrt{17})}{4}=\frac{6+18+(\sqrt{17}-\sqrt{17})}{4}=\frac{6+18}{4}$

Agree ?

By the way, for the RHS : $(9-\sqrt{17})(9+\sqrt{17})=9^2-(\sqrt{17})^2=81-17$
thanks to the difference of squares formula : $(a-b)(a+b)=a^2-b^2$

I'm looking for another method of solving it, not sure I'll find one >.<

**j0k3r** · September 22nd 2008, 11:41 AM

Originally Posted by Moo

Hmmm no !

$\frac 32+\frac{9-\sqrt{17}}{4}+\frac{9+\sqrt{17}}{4}=\frac 64+\frac{9-\sqrt{17}}{4}+\frac{9+\sqrt{17}}{4}$

Getting everything on a unique fraction :

$=\frac{6+(9-\sqrt{17})+(9+\sqrt{17})}{4}=\frac{6+18+(\sqrt{17}-\sqrt{17})}{4}=\frac{6+18}{4}$

Agree ?

By the way, for the RHS : $(9-\sqrt{17})(9+\sqrt{17})=9^2-(\sqrt{17})^2=81-17$
thanks to the difference of squares formula : $(a-b)(a+b)=a^2-b^2$

I'm looking for another method of solving it, not sure I'll find one >.<

$\frac{6}{4} + \frac{9-\sqrt{17}}{4} + \frac{9+\sqrt{17}}{4} = \frac{6}{4}*\frac{9-\sqrt{17}}{4}*\frac{9+\sqrt{17}}{4}$ .

I GOT IT

Solution for those that want it:

$6+(9-\sqrt{17})+(9+\sqrt{17})=\frac{3}{2}*\frac{9-\sqrt{17}}{4}*\frac{9+\sqrt{17}}{4}*4$

$6+18=\frac{3(9-\sqrt{17})}{4*2}*({9+\sqrt{17})}$

$8*24=3(81-17)$

$192=3(64)$

$192=192$

Hell yeh

Mo.

**Soroban** · September 22nd 2008, 11:57 AM

Hello, j0k3r!

Are you really having difficulty with Arithmetic?

Given: . $a \:= \:\frac{3}{2},\;\;b \:=\: \frac{9-\sqrt{17}}{4},\;\;c \:= \:\frac{9+\sqrt{17}}{4}$

Show that: . $a + b + c \:= \:abc$ .

$a + b + c \:=\:\frac{3}{2} + \underbrace{\frac{9-\sqrt{17}}{4} + \frac{9 + \sqrt{17}}{4}}$

. . . . . . $= \;\frac{3}{2} + \overbrace{\frac{9 - \sqrt{17} + 9 + \sqrt{17}}{4}}$

. . . . . . $= \;\frac{3}{2} + \frac{18}{4}$

. . . . . . $=\;\frac{3}{2} + \frac{9}{2}$

. . . . . . $=\;\frac{12}{2}$

. . . . . . $=\;{\color{blue}6}$

$a\!\cdot\!b\!\cdot\!c \;=\;\left(\frac{3}{2}\right)\underbrace{\left(\fr ac{9-\sqrt{17}}{4}\right)\left(\frac{9+\sqrt{17}}{4}\ri ght)}$

. . . . $= \;\left(\frac{3}{2}\right)\overbrace{\left(\frac{8 1 + 9\sqrt{17} - 9\sqrt{17} - 17}{16}\right)}$

. . . . $= \;\frac{3}{2}\cdot\frac{64}{16}$

. . . . $=\;{\color{blue}6}$

**j0k3r** · September 22nd 2008, 12:04 PM

Originally Posted by Soroban

Hello, j0k3r!

Are you really having difficulty with Arithmetic?

Uhm no... Did you bother to read the fact that I had solved it? o_O.

Mo.

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