- Thread starter
- #1
Albert
Well-known member
- Jan 25, 2013
- 1,225
ab=1
bc=2
cd=3
de=4
ea=5
find a,b,c,d,e
bc=2
cd=3
de=4
ea=5
find a,b,c,d,e
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: $ade^2=4(5)$ $2ad(e^2)=2(20)$ $3(e^2)=40$ $\therefore e=\pm 2\sqrt{\dfrac{10}{3}}$ | $\begin{align*}\therefore a&=\dfrac{5}{e}\\&=\pm \dfrac{5}{2}\sqrt{\dfrac{3}{10}}\end{align*}$ $\begin{align*}\therefore d&=\dfrac{3}{2a}\\&=\pm \dfrac{3}{5}\sqrt{\dfrac{10}{3}}\end{align*}$ $\begin{align*}\therefore c&=\dfrac{3}{d}\\&=\pm 5 \sqrt{\dfrac{3}{10}}\end{align*}$ $\begin{align*}\therefore b&=\dfrac{1}{a}\\&=\pm \dfrac{2}{5}\sqrt{\dfrac{10}{3}}\end{align*}$ |
My solution:
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*}$
$\begin{align*}\therefore d&=\dfrac{3}{2a}\\&=\pm \dfrac{3}{5}\sqrt{\dfrac{10}{3}}\end{align*}$
$\begin{align*}\therefore c&=\dfrac{3}{d}\\&=\pm 5 \sqrt{\dfrac{3}{10}}\end{align*}$
$\begin{align*}\therefore b&=\dfrac{1}{a}\\&=\pm \dfrac{2}{5}\sqrt{\dfrac{10}{3}}\end{align*}$
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
The \sqrt{} command creates a square root surrounding an expression.As I do not know how to put square root I have put power 1/2
Well done, kali!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 solutionAs 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