contestada

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.

Respuesta :

Answer:

[tex]\displaystyle\frac{3\sqrt{3}}{4}[/tex]

Explanation:

Given,

& is defined,

[tex]a \& b = \displaystyle\frac{\sqrt{a b + a}}{\sqrt{a b - b}}[/tex]

If a = 9, b = 2,

[tex]9 \& 2 = \displaystyle\frac{\sqrt{(9)(2)+ (9)}}{\sqrt{(9)(2)- 2}}[/tex]

[tex] = \displaystyle\frac{\sqrt{18 + 9}}{\sqrt{18 - 2}}[/tex]

[tex]= \displaystyle\frac{\sqrt{27}}{\sqrt{16}}[/tex]

[tex]=\displaystyle\frac{\sqrt{9}\sqrt{3}}{4}[/tex]  ( ∵√16 = 4 )

[tex]=\displaystyle\frac{3\sqrt{3}}{4}[/tex]     ( ∵ √9 = 3 )

∵ Further simplification is not possible,

Hence, the required simplified form is,

[tex]\displaystyle\frac{3\sqrt{3}}{4}[/tex]