Answer:
3√3
Step-by-step explanation:
For the problem shown here, your answer 3√3 is correct.
When there is a radical by itself in the denominator, you multiply numerator and denominator by a radical that results in the product being rational. For a square root, that will usually be the same square root:
[tex]\dfrac{9}{\sqrt{3}}=\dfrac{9}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{9\sqrt{3}}{3}=\boxed{3\sqrt{3}}[/tex]
__
If the problem has a sum in the denominator involving a square root, then you multiply numerator and denominator by the conjugate of that sum (the sum with the sign changed). This uses the special product "difference of squares" to eliminate the radical term.
Example:
[tex]\dfrac{9}{2-\sqrt{3}}=\dfrac{9}{2-\sqrt{3}}\cdot\dfrac{2+\sqrt{3}}{2+\sqrt{3}}=\dfrac{9(2+\sqrt{3})}{2^2-(\sqrt{3})^2}=\dfrac{18+9\sqrt{3}}{4-3}\\\\=\boxed{18+9\sqrt{3}}[/tex]
__
It is easy to demonstrate that none of the offered choices for this problem has the same value as 9/√3.
9/√3 ≈ 5.196. Offered choices have values of about 4.798, 1.732, 6.681, 23.196 -- none even close.
Please discuss this question with your teacher.
