Question

Find the mean proportional between 6+√27 and 8-√48 in its simplest form​

Answer 1

$\textbf{Given:}$

$\text{ 6+\sqrt{27} and 8-\sqrt{48} }$

$\textbf{To find:}$

$\text{The mean proportional between the given two numbers}$

$\textbf{Solution:}$

$\textbf{Concept:}$

$\text{The mean proportional between the two numbers a and b}$

$\text{is \sqrt{ab} }$

$\text{Now,}$

$\textbf{The mean proportional between 6+\sqrt{27} and 8-\sqrt{48} }$

$=\sqrt{(6+\sqrt{27})(8-\sqrt{48})}$

$=\sqrt{(6+\sqrt{9{\times}3})(8-\sqrt{16{\times}3})}$

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

$=\sqrt{3(2+\sqrt{3}){\times}4(2-\sqrt{3})}$

$=\sqrt{12(2+\sqrt{3}){\times}(2-\sqrt{3})}$

$\text{Using the idenitity}$

$\boxed{(a-b)(a+b)=a^2-b^2}$

$=\sqrt{12(2^2-\sqrt{3}^2)}$

$=\sqrt{12(4-3)}$

$=\sqrt{12(1)}$

$=\sqrt{12}$

$=\sqrt{4{\times}3}$

$=2\sqrt{3}$

$\textbf{Answer:}$

$\textbf{The mean proportional between the given two numbers is \bf\,2\sqrt{3} }$.

Answer 2

Answer:

Given:

$6 \: + \: \sqrt{27} \: and \: \: 8 \: - \: \sqrt{48}$

The mean proportional between the given two numbers.

Solution:

Concept:

The mean proportional between the two numbers a and b is

$\sqrt{ab}$

Now,

The mean proportional between

$6 \: + \: \sqrt{27} \: and \: 8 - \sqrt{48}$

$\implies \sqrt{(6 + \sqrt{27})(8 - \sqrt{48} )} \\ \implies \: \sqrt{(6 + \sqrt{9 \times 3} (8 - \sqrt{16 \times 3} )} \\ \implies \: \sqrt{(6 + 3 \sqrt{3})(8 - 4 \sqrt{3} ) } \\ \implies \: \sqrt{3(2 + \sqrt{3}) \times 4(2 - \sqrt{3} )} \\ \implies \: \sqrt{12(2 + \sqrt{3} ) \times (2 - \sqrt{3} } \\ \\ \huge \color {red} \mathfrak using \: the \: idenitity \\ \\ \huge \underline \color {green} ( a - b)(a + b) = {a}^{2} - {b}^{2}$

$\implies \: \sqrt{12( {2}^{2 - \: \sqrt{ {3}^{2} } })} \\ \implies \: \sqrt{2(4 - 3)} \: \\ \implies \: \sqrt{12(1)} \\ \implies \: \sqrt{12} \\ \implies \: \sqrt{4 \times 3} \\ \implies \huge \color {red} {2 \sqrt{3}} \\ \\ \small \mathcal the \: mean \: proportional \: between \: \\ the \: given \: two \: number \: is \: 2 \sqrt{3} .$