Question

select all the expressions that are irrational.
show your work here
-√3 + 2 √2·√3
√27·√3 3√7 + 2√7 - 5√7
√6 + 3√2

Explanation:

Step1: Recall irrational number definition

An irrational number is a number that cannot be expressed as a fraction of two integers. For expressions with radicals, simplify and check if the result is irrational (non - repeating, non - terminating decimal or cannot be expressed as a fraction).

Step2: Analyze \(-\sqrt{3}+2\)

\(\sqrt{3}\) is irrational (since \(3\) is not a perfect square). The sum of an irrational number (\(-\sqrt{3}\)) and a rational number (\(2\)) is irrational. So \(-\sqrt{3}+2\) is irrational.

Step3: Analyze \(\sqrt{2}\cdot\sqrt{3}\)

Using the property \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\) (\(a\geq0,b\geq0\)), we have \(\sqrt{2}\cdot\sqrt{3}=\sqrt{2\times3}=\sqrt{6}\). Since \(6\) is not a perfect square, \(\sqrt{6}\) is irrational. So \(\sqrt{2}\cdot\sqrt{3}\) is irrational.

Step4: Analyze \(\sqrt{27}\cdot\sqrt{3}\)

First, simplify \(\sqrt{27}=\sqrt{9\times3} = 3\sqrt{3}\). Then \(\sqrt{27}\cdot\sqrt{3}=3\sqrt{3}\cdot\sqrt{3}=3\times3 = 9\), which is a rational number (can be written as \(\frac{9}{1}\)).

Step5: Analyze \(3\sqrt{7}+2\sqrt{7}-5\sqrt{7}\)

Combine like terms: \((3 + 2-5)\sqrt{7}=0\times\sqrt{7}=0\), which is a rational number (can be written as \(\frac{0}{1}\)).

Step6: Analyze \(\sqrt{6}+3\sqrt{2}\)

\(\sqrt{6}\) and \(\sqrt{2}\) are both irrational. The sum of two irrational numbers (after checking for like terms, here \(\sqrt{6}\) and \(\sqrt{2}\) are not like radicals and cannot be combined to a rational number) is irrational. So \(\sqrt{6}+3\sqrt{2}\) is irrational.

Answer:

\(-\sqrt{3}+2\), \(\sqrt{2}\cdot\sqrt{3}\), \(\sqrt{6}+3\sqrt{2}\)