Question

What should come in place x in the equation $\frac{x}{\sqrt{128}}=\frac{\sqrt{162}}{x}$

Answer 1

Let $\frac{x}{\sqrt{128}}=\frac{\sqrt{162}}{x}$
Then $x^2=\sqrt{128\times 162}$
$=\sqrt{64\times 2\times 18\times 9}=\sqrt{8^2\times 6^2\times 3^2}=8\times 6\times 3=144\therefore x=\sqrt{144}=12$