Question

which of the following is a non-real complex number?
a. $3\sqrt{\frac{7}{5}} + \sqrt{-\frac{9}{5}}$
b. $\frac{5 - 2\sqrt{7}}{4}$
c. $9 - \frac{1}{\sqrt{17}}$
d. $-8\sqrt{\frac{1}{6}} - \frac{3}{\sqrt{2}}$

Explanation:

Step1: Recall complex number form

A complex number is in the form \(a + bi\), where \(a,b\in\mathbb{R}\) and \(i=\sqrt{-1}\). A non - real complex number has \(b
eq0\) (i.e., it has an imaginary part). For real numbers, the imaginary part \(b = 0\) (no square root of a negative number).

Step2: Analyze Option A

For \(3\sqrt{\frac{7}{5}}+\sqrt{-\frac{9}{5}}\), we can rewrite \(\sqrt{-\frac{9}{5}}\) using the definition of the imaginary unit \(i=\sqrt{-1}\). So \(\sqrt{-\frac{9}{5}}=\sqrt{\frac{9}{5}\times(- 1)}=\sqrt{\frac{9}{5}}\times\sqrt{-1}=\frac{3}{\sqrt{5}}i\). Then the number becomes \(3\sqrt{\frac{7}{5}}+\frac{3}{\sqrt{5}}i\), which has a non - zero imaginary part (\(b = \frac{3}{\sqrt{5}}
eq0\)), so it is a non - real complex number.

Step3: Analyze Option B

\(\frac{5 - 2\sqrt{7}}{4}=\frac{5}{4}-\frac{2\sqrt{7}}{4}\), this is a real number (it can be written as \(a+bi\) with \(b = 0\)).

Step4: Analyze Option C

\(9-\frac{1}{\sqrt{17}}\) is a real number (it can be written as \(a + bi\) with \(b=0\)).

Step5: Analyze Option D

Simplify \(-8\sqrt{\frac{1}{6}}-\frac{3}{\sqrt{2}}\):
First, \(-8\sqrt{\frac{1}{6}}=-\frac{8}{\sqrt{6}}=-\frac{4\sqrt{6}}{3}\) (rationalizing the denominator: \(\frac{8}{\sqrt{6}}=\frac{8\sqrt{6}}{6}=\frac{4\sqrt{6}}{3}\)), and \(\frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2}\) (rationalizing the denominator). The number \(-\frac{4\sqrt{6}}{3}-\frac{3\sqrt{2}}{2}\) is a real number (it can be written as \(a + bi\) with \(b = 0\)).

Answer:

A. \(3\sqrt{\frac{7}{5}}+\sqrt{-\frac{9}{5}}\)