Question

Find the mean proportional between :

(i) $6+3\sqrt{3}$ and $8-4\sqrt{3}$

(ii) a - b and $a^3-a^{2}b.$

Answer 1

(i) Let the mean proportional between $6+3\sqrt{3}and8-4\sqrt{3}bex$
$6+3\sqrt{3}.xand8-4\sqrt{3}$ are in continued proportion.

$6+3\sqrt{3}:x=x:8-4\surd 3$
⟹ $x\times x=(6+3\sqrt{3})(8-4\sqrt{3})$
⟹$x^2=48+24\sqrt{3}-24\sqrt{3}-36$
⟹ $x^2=12$
⟹ $x=2\sqrt{3}$
(ii) Let the mean proportional between a − b and $a^3-a^{2}b$ be x.
⟹ a − b, x, $a^3-a^{2}b$ are in continued proportion.
⟹ a − b : x = x : $a^3-a^{2}b$

⟹$x\times x=(a-b)(a^3-a^{2}b)$

⟹ $x^2=(a-b)\left[a^2\right(a-b\left)\right]$

⟹ $x^2=(a-b)(a-b)\left(a^2\right)$

⟹ $x^2=(a-b{)}^2\times (a^2)$
⟹ ${\sqrt{x}}^2=\sqrt{(}a-b)a^2$
⟹ x = a (a − b)