Question

5
select all the correct locations on the table.
select all of the non - real complex numbers in the given table.
$5 - \sqrt{\frac{9}{4}}$
$1 + \sqrt{-3}$
$4 - 3\sqrt{-16}$
$\frac{2 - \sqrt{12}}{5}$
$\frac{3 + 2\sqrt{-9}}{7}$
$9\sqrt{-\frac{7}{5}}$

Explanation:

Step1: Recall non - real complex numbers

A complex number \(a + bi\) is non - real if \(b
eq0\) (or in the case of square roots, if the radicand of a square root is negative, since \(\sqrt{-c}=i\sqrt{c}\) for \(c>0\)).

Step2: Analyze \(5-\sqrt{\frac{9}{4}}\)

\(\sqrt{\frac{9}{4}}=\frac{3}{2}\), so \(5 - \sqrt{\frac{9}{4}}=5-\frac{3}{2}=\frac{10 - 3}{2}=\frac{7}{2}\), which is a real number.

Step3: Analyze \(1+\sqrt{-3}\)

Since the radicand \(- 3<0\), \(\sqrt{-3}=i\sqrt{3}\), so \(1+\sqrt{-3}=1 + i\sqrt{3}\), which is a non - real complex number.

Step4: Analyze \(4-3\sqrt{-16}\)

\(\sqrt{-16}=i\sqrt{16} = 4i\), so \(4-3\sqrt{-16}=4-3\times4i=4 - 12i\), which is a non - real complex number.

Step5: Analyze \(\frac{2-\sqrt{12}}{5}\)

\(\sqrt{12}=2\sqrt{3}\), so \(\frac{2-\sqrt{12}}{5}=\frac{2 - 2\sqrt{3}}{5}\), which is a real number.

Step6: Analyze \(\frac{3 + 2\sqrt{-9}}{7}\)

\(\sqrt{-9}=i\sqrt{9}=3i\), so \(\frac{3+2\sqrt{-9}}{7}=\frac{3 + 2\times3i}{7}=\frac{3+6i}{7}=\frac{3}{7}+\frac{6}{7}i\), which is a non - real complex number.

Step7: Analyze \(9\sqrt{-\frac{7}{5}}\)

Since the radicand \(-\frac{7}{5}<0\), \(\sqrt{-\frac{7}{5}}=i\sqrt{\frac{7}{5}}\), so \(9\sqrt{-\frac{7}{5}}=9i\sqrt{\frac{7}{5}}\), which is a non - real complex number.

Answer:

The non - real complex numbers are \(1+\sqrt{-3}\), \(4 - 3\sqrt{-16}\), \(\frac{3 + 2\sqrt{-9}}{7}\), \(9\sqrt{-\frac{7}{5}}\) (i.e., the second, third, fifth, and sixth entries in the table).