find a,b,c,d,e

[Albert]

ab=1
bc=2
cd=3
de=4
ea=5
find a,b,c,d,e

 

[anemone]

My solution:

Spoiler

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 $c=2a$.

From $cd=3$ and $c=2a$, we have: $(2a)d=3$ $2ad=3$

From $de=4$ and $ea=5$ and $2ad=3$, we have: $ade^2=4(5)$ $2ad(e^2)=2(20)$ $3(e^2)=40$ $\therefore e=\pm 2\sqrt{\dfrac{10}{3}}$

$\begin{align*}\therefore a&=\dfrac{5}{e}\\&=\pm \dfrac{5}{2}\sqrt{\dfrac{3}{10}}\end{align*}$ $\begin{align*}\therefore d&=\dfrac{3}{2a}\\&=\pm \dfrac{3}{5}\sqrt{\dfrac{10}{3}}\end{align*}$ $\begin{align*}\therefore c&=\dfrac{3}{d}\\&=\pm 5 \sqrt{\dfrac{3}{10}}\end{align*}$ $\begin{align*}\therefore b&=\dfrac{1}{a}\\&=\pm \dfrac{2}{5}\sqrt{\dfrac{10}{3}}\end{align*}$

 

[kaliprasad]

As I do not know how to put square root I have put power 1/2

Spoiler

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

 

[kaliprasad]

My solution:

Spoiler

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 $c=2a$.

From $cd=3$ and $c=2a$, we have: $(2a)d=3$ $2ad=3$

From $de=4$ and $ea=5$ and $2ad=3$, we have: $ade^2=4(5)$ $2ad(e^2)=2(20)$ $3(e^2)=40$ $\therefore e=\pm 2\sqrt{\dfrac{10}{3}}$

$\begin{align*}\therefore a&=\dfrac{5}{e}\\&=\pm \dfrac{5}{2}\sqrt{\dfrac{3}{10}}\end{align*}$ $\begin{align*}\therefore d&=\dfrac{3}{2a}\\&=\pm \dfrac{3}{5}\sqrt{\dfrac{10}{3}}\end{align*}$ $\begin{align*}\therefore c&=\dfrac{3}{d}\\&=\pm 5 \sqrt{\dfrac{3}{10}}\end{align*}$ $\begin{align*}\therefore b&=\dfrac{1}{a}\\&=\pm \dfrac{2}{5}\sqrt{\dfrac{10}{3}}\end{align*}$

Spoiler

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

 

[MarkFL]

Spoiler

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

Spoiler

Yes, you can see that all of the values anemone found inherit their sign from $e$.

 

[anemone]

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 $\sqrt{x}$.

Spoiler

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! And looking more closely, I'd say we're actually approached the problem using quite similar concept!

 

[Albert]

As I do not know how to put square root I have put power 1/2

Spoiler

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

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


you will get: $\sqrt[m]{b^n}$