Question

The operation $\&$ is defined for positive integers $a$ and $b$ as $a \& b = \displaystyle\frac{\sqrt{a b + a}}{\sqrt{a b - b}}$. What is the value of $9 \& 2$? Express your answer as a common fraction in simplest radical form.

Answer 1

`\displaystyle(3√(3))/(4)`

Step-by-step explanation:

Given,

& is defined,

`a \& b = \displaystyle(√(a b + a))/(√(a b - b))`

If a = 9, b = 2,

`9 \& 2 = \displaystyle(√((9)(2)+ (9)))/(√((9)(2)- 2))`

`= \displaystyle(√(18 + 9))/(√(18 - 2))`

`= \displaystyle(√(27))/(√(16))`

`=\displaystyle(√(9)√(3))/(4)` ( ∵√16 = 4 )

`=\displaystyle(3√(3))/(4)` ( ∵ √9 = 3 )

∵ Further simplification is not possible,

Hence, the required simplified form is,

`\displaystyle(3√(3))/(4)`