Answer:
[tex]18\sqrt2[/tex]
Step-by-step explanation:
To simplify:
[tex]2 \sqrt{18}+ 3 \sqrt2+ \sqrt{162 }[/tex]
First of all, let us write 18 and 162 as product of prime factors:
[tex]18 = 2 \times \underline{3 \times 3}\\162 = 2 \times \underline{3 \times 3} \times \underline{3 \times 3}[/tex]
The pairs are underlined as above.
While taking roots, only one of the numbers from the pairs will be chosen.
Now, taking square roots.
[tex]\sqrt{18} =3 \sqrt2[/tex]
[tex]162 = 3 \times 3 \times \sqrt 2 = 9 \sqrt2[/tex]
So, the given expression becomes:
[tex]2 \sqrt{18}+ 3 \sqrt2+ \sqrt{162 } = 2 \times 3\sqrt2 + 3\sqrt2 +9\sqrt2\\\Rightarrow 6\sqrt2 + 3\sqrt2 +9\sqrt2\\\Rightarrow \sqrt2(6+3+9)\\\Rightarrow \bold{18\sqrt2}[/tex]
So, the answer is:
[tex]18\sqrt2[/tex] or 18 StartRoot 2 EndRoot
