53.3k views
5 votes
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.

1 Answer

0 votes

Answer:


\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)

User JPvdMerwe
by
7.6k points