Question

question 6.
find the sum:
\\(\sqrt{18} + \sqrt{12}\\)
a. \\(5\sqrt{6}\\)
b. \\(3\sqrt{2} + 2\sqrt{3}\\)
c. \\(6\sqrt{5}\\)
d. \\(9\sqrt{2} + 4\sqrt{3}\\)

Explanation:

Step1: Simplify $\sqrt{18}$

Factor 18 into $9\times2$, so $\sqrt{18}=\sqrt{9\times2}$. Using the property $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ ($a\geq0,b\geq0$), we get $\sqrt{9\times2}=\sqrt{9}\times\sqrt{2}=3\sqrt{2}$.

Step2: Simplify $\sqrt{12}$

Factor 12 into $4\times3$, so $\sqrt{12}=\sqrt{4\times3}$. Using the same property, we get $\sqrt{4\times3}=\sqrt{4}\times\sqrt{3}=2\sqrt{3}$.

Step3: Add the simplified terms

Now, $\sqrt{18}+\sqrt{12}=3\sqrt{2}+2\sqrt{3}$.

Answer:

B. $3\sqrt{2}+2\sqrt{3}$