MHB Find the Values of a,b,c,d,e (ab=1, bc=2, cd=3, de=4, ea=5)

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The discussion focuses on solving a system of equations defined by the products ab=1, bc=2, cd=3, de=4, and ea=5 to find the values of a, b, c, d, and e. Participants share their solutions and methods, with one user mentioning the use of square roots and expressing difficulty in formatting them correctly. The conversation highlights the importance of using the correct mathematical notation, specifically the square root command. Overall, the thread emphasizes collaborative problem-solving and the sharing of mathematical techniques. The users demonstrate a supportive environment for tackling complex equations.
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ab=1
bc=2
cd=3
de=4
ea=5
find a,b,c,d,e
 
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My solution:
[TABLE="class: grid, width: 800"]
[TR]
[TD]From $ab=1$ and $bc=2$, we have:

$2ab=bc$

$2ab-bc=0$

$b(2a-c)=0$

Since $b \ne 0$, $2a-c=0$ must be true or $c=2a$.[/TD]
[TD]From $cd=3$ and $c=2a$, we have:

$(2a)d=3$

$2ad=3$[/TD]
[TD]From $de=4$ and $ea=5$ and $2ad=3$, we have:

$ade^2=4(5)$

$2ad(e^2)=2(20)$

$3(e^2)=40$

$\therefore e=\pm 2\sqrt{\dfrac{10}{3}}$[/TD]
[TD]$\begin{align*}\therefore a&=\dfrac{5}{e}\\&=\pm \dfrac{5}{2}\sqrt{\dfrac{3}{10}}\end{align*}$

$\begin{align*}\therefore d&=\dfrac{3}{2a}\\&=\pm \dfrac{3}{5}\sqrt{\dfrac{10}{3}}\end{align*}$

$\begin{align*}\therefore c&=\dfrac{3}{d}\\&=\pm 5 \sqrt{\dfrac{3}{10}}\end{align*}$

$\begin{align*}\therefore b&=\dfrac{1}{a}\\&=\pm \dfrac{2}{5}\sqrt{\dfrac{10}{3}}\end{align*}$
[/TD]
[/TR]
[/TABLE]
 
As I do not know how to put square root I have put power 1/2

Multiply all 5 to get (abcde)$^2$ = 120
Or abcde = +/-120$^{(1/2)}$
Devide by product of ab and cd to get e = = +/-120$^{(1/2)}$/ 3= = +/-(40/3)$^{(1/2})$ = +/-2(10/3)$^{(1/2)}$

Similarly you can find the rest
 
anemone said:
My solution:
[TABLE="class: grid, width: 800"]
[TR]
[TD]From $ab=1$ and $bc=2$, we have:

$2ab=bc$

$2ab-bc=0$

$b(2a-c)=0$

Since $b \ne 0$, $2a-c=0$ must be true or $c=2a$.[/TD]
[TD]From $cd=3$ and $c=2a$, we have:

$(2a)d=3$

$2ad=3$[/TD]
[TD]From $de=4$ and $ea=5$ and $2ad=3$, we have:

$ade^2=4(5)$

$2ad(e^2)=2(20)$

$3(e^2)=40$

$\therefore e=\pm 2\sqrt{\dfrac{10}{3}}$[/TD]
[TD]$\begin{align*}\therefore a&=\dfrac{5}{e}\\&=\pm \dfrac{5}{2}\sqrt{\dfrac{3}{10}}\end{align*}$

$\begin{align*}\therefore d&=\dfrac{3}{2a}\\&=\pm \dfrac{3}{5}\sqrt{\dfrac{10}{3}}\end{align*}$

$\begin{align*}\therefore c&=\dfrac{3}{d}\\&=\pm 5 \sqrt{\dfrac{3}{10}}\end{align*}$

$\begin{align*}\therefore b&=\dfrac{1}{a}\\&=\pm \dfrac{2}{5}\sqrt{\dfrac{10}{3}}\end{align*}$
[/TD]
[/TR]
[/TABLE]
I thought that it may be noted that all are positive or all are -ve. I know you know it but for benefit of others
 
kaliprasad said:
I thought that it may be noted that all are positive or all are -ve. I know you know it but for benefit of others

Yes, you can see that all of the values anemone found inherit their sign from $e$. :D
 
kaliprasad said:
As I do not know how to put square root I have put power 1/2

The \sqrt{} command creates a square root surrounding an expression.

Take for example, \sqrt{x} gives $\sqrt{x}$.

kaliprasad said:
Multiply all 5 to get (abcde)$^2$ = 120
Or abcde = +/-120$^{(1/2)}$
Devide by product of ab and cd to get e = = +/-120$^{(1/2)}$/ 3= = +/-(40/3)$^{(1/2})$ = +/-2(10/3)$^{(1/2)}$

Similarly you can find the rest

Well done, kali!(Sun) And looking more closely, I'd say we're actually approached the problem using quite similar concept!:o
 
kaliprasad said:
As I do not know how to put square root I have put power 1/2

Multiply all 5 to get (abcde)$^2$ = 120
Or abcde = +/-120$^{(1/2)}$
Devide by product of ab and cd to get e = = +/-120$^{(1/2)}$/ 3= = +/-(40/3)$^{(1/2})$ = +/-2(10/3)$^{(1/2)}$

Similarly you can find the rest
good solution (Clapping)

the use of square root example :type :" \sqrt[m]{b^n} " between two "dollar signals" you will get: $\sqrt[m]{b^n}$
 

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