Simplification

**lds09** · Apr 3rd 2008, 09:38 PM

Hello,

If $\frac{a}{b} = \frac{b}{c} = \frac{c}{d},$ prove that
$\frac{a}{d}$ is equal to $\sqrt{\frac{a^5 + b^2c^2 + a^3c^2}{b^4c + d^4 + b^2cd^2}}$

It seems like a mess. Any suggestions on where to start?

Btw, this problem is from Higher Algebra by H.S Hall and S. R Knight, a textbook that is still used in India even though it was written in the 1800's.

**JaneBennet** · Apr 3rd 2008, 11:04 PM

We are assuming that $a,b,c,d>0$ , right? (If not, you could have, say, $a=c=1$ and $b=d=-1$ and then $\frac{a}{d}$ would be negative.)

Looking at the expression, the best thing to do first looks like eliminating $b^2=ac$ :

$\color{white}.\quad.$ $\frac{a^5+b^2c^2+a^3c^2}{b^4c+d^4+b^2cd^2}$

$=\ \frac{a^5+ac^3+a^3c^2}{a^2c^3+d^4+ac^2d^2}$

We also have $a^2d^2=b^2c^2=ac^3$ , i.e. $c^3=ad^2$ . So

$\color{white}.\quad.$ $\frac{a^5+ac^3+a^3c^2}{a^2c^3+d^4+ac^2d^2}$

$=\ \frac{a^5+a^2d^2+a^3c^2}{a^3d^2+d^4+ac^2d^2}$

$=\ \frac{a^2\left(a^3+d^2+ac^2\right)}{d^2\left(a^3+d ^2+ac^2\right)}$

$=\ \frac{a^2}{d^2}$

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