Question

Find the third proportional to:
$a^2-5a+6,a^2+a-6,a^2-9$

Answer 1

(a)Third proportional to $\left(a^2-5a+6\right),\left(a^2+a-6\right)$

= Third proportional to $\left(a-2\right)\left(a-3\right),\left(a-2\right)\left(a+3\right)$

Let the third proportional be p. Then:

$\frac{\left(a-2\right)\left(a-3\right)}{\left(a-2\right)\left(a+3\right)}=\frac{\left(a-2\right)\left(a+3\right)}{p}$

$\Rightarrow p=\frac{{\left[\left(a-2\right)\left(a+3\right)\right]}^2}{\left(a-2\right)\left(a-3\right)}$

$\Rightarrow p=\left(a-2\right)\frac{{\left(a+3\right)}^2}{a-3}$

(b) Find the Fourth proportional to

$a^2-5a+6,a^2+a-6,a^2-9$

Let the fourth proportional be p. Then:

$\frac{a^2-5a+6}{a^2+a-6}=\frac{a^2-9}{p}$

$\Rightarrow \frac{\left(a-2\right)\left(a-3\right)}{\left(a-2\right)\left(a+3\right)}=\frac{\left(a-3\right)\left(a+3\right)}{p}$

$\Rightarrow p=\frac{\left(a-2\right)\left(a+3\right)\left(a-3\right)\left(a+3\right)}{\left(a-2\right)\left(a-3\right)}$

$\Rightarrow p={\left(a+3\right)}^2$

So fourth proportional is ${\left(a+3\right)}^2$