Answer :
Certainly! Let's go through each of the given statements to determine which ones are true.
### A. [tex]\(\sqrt{-2} = \sqrt{2}\)[/tex]
This statement is false. The square root of [tex]\(-2\)[/tex] involves an imaginary number. Specifically, [tex]\(\sqrt{-2} = i\sqrt{2}\)[/tex], where [tex]\(i\)[/tex] is the imaginary unit ([tex]\(i^2 = -1\)[/tex]). Therefore, [tex]\(\sqrt{-2}\)[/tex] is not equal to the real number [tex]\(\sqrt{2}\)[/tex].
### B. [tex]\(\sqrt{-2} = \sqrt{2i}\)[/tex]
This statement is also false. Again, [tex]\(\sqrt{-2} = i\sqrt{2}\)[/tex], and [tex]\(\sqrt{2i}\)[/tex] is a different complex number. To evaluate [tex]\(\sqrt{2i}\)[/tex]:
- Convert [tex]\(2i\)[/tex] into polar form: [tex]\(2i = 2e^{i\pi/2}\)[/tex].
- Taking the square root: [tex]\(\sqrt{2i} = \sqrt{2 \cdot e^{i\pi/2}} = \sqrt{2} \cdot e^{i\pi/4} = \sqrt{2} \cdot \left(\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\right) = 1 + i\)[/tex].
Clearly, [tex]\(1 + i\)[/tex] is not equal to [tex]\(i\sqrt{2}\)[/tex]. Therefore, [tex]\(\sqrt{-2} \neq \sqrt{2i}\)[/tex].
### C. [tex]\(\sqrt{-8} = 2i\sqrt{2}\)[/tex]
This statement is true. Let's simplify [tex]\(\sqrt{-8}\)[/tex]:
[tex]\[ \sqrt{-8} = \sqrt{8 \cdot -1} = \sqrt{8} \cdot \sqrt{-1} = 2\sqrt{2} \cdot i = 2i\sqrt{2}. \][/tex]
Thus, [tex]\(\sqrt{-8} = 2i\sqrt{2}\)[/tex] is indeed correct.
### D. [tex]\(\sqrt{-8} = -2\sqrt{2}\)[/tex]
This statement is false. As seen above, [tex]\(\sqrt{-8} = 2i\sqrt{2}\)[/tex]. Since the square root of a negative number involves an imaginary unit [tex]\(i\)[/tex], [tex]\(\sqrt{-8}\)[/tex] cannot be a real number like [tex]\(-2\sqrt{2}\)[/tex].
### E. [tex]\(\sqrt{-16} = 4i\)[/tex]
This statement is true. Let's simplify [tex]\(\sqrt{-16}\)[/tex]:
[tex]\[ \sqrt{-16} = \sqrt{16 \cdot -1} = \sqrt{16} \cdot \sqrt{-1} = 4 \cdot i = 4i. \][/tex]
Thus, [tex]\(\sqrt{-16} = 4i\)[/tex] is correct.
### F. [tex]\(\sqrt{-16} = -4\)[/tex]
This statement is false. As established, [tex]\(\sqrt{-16} = 4i\)[/tex], which involves an imaginary unit. Therefore, [tex]\(\sqrt{-16}\)[/tex] cannot equal [tex]\(-4\)[/tex], a real number.
### Conclusion
The true statements are:
- C. [tex]\(\sqrt{-8} = 2i\sqrt{2}\)[/tex]
- E. [tex]\(\sqrt{-16} = 4i\)[/tex]
Therefore, the true statements are C and E.
### A. [tex]\(\sqrt{-2} = \sqrt{2}\)[/tex]
This statement is false. The square root of [tex]\(-2\)[/tex] involves an imaginary number. Specifically, [tex]\(\sqrt{-2} = i\sqrt{2}\)[/tex], where [tex]\(i\)[/tex] is the imaginary unit ([tex]\(i^2 = -1\)[/tex]). Therefore, [tex]\(\sqrt{-2}\)[/tex] is not equal to the real number [tex]\(\sqrt{2}\)[/tex].
### B. [tex]\(\sqrt{-2} = \sqrt{2i}\)[/tex]
This statement is also false. Again, [tex]\(\sqrt{-2} = i\sqrt{2}\)[/tex], and [tex]\(\sqrt{2i}\)[/tex] is a different complex number. To evaluate [tex]\(\sqrt{2i}\)[/tex]:
- Convert [tex]\(2i\)[/tex] into polar form: [tex]\(2i = 2e^{i\pi/2}\)[/tex].
- Taking the square root: [tex]\(\sqrt{2i} = \sqrt{2 \cdot e^{i\pi/2}} = \sqrt{2} \cdot e^{i\pi/4} = \sqrt{2} \cdot \left(\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\right) = 1 + i\)[/tex].
Clearly, [tex]\(1 + i\)[/tex] is not equal to [tex]\(i\sqrt{2}\)[/tex]. Therefore, [tex]\(\sqrt{-2} \neq \sqrt{2i}\)[/tex].
### C. [tex]\(\sqrt{-8} = 2i\sqrt{2}\)[/tex]
This statement is true. Let's simplify [tex]\(\sqrt{-8}\)[/tex]:
[tex]\[ \sqrt{-8} = \sqrt{8 \cdot -1} = \sqrt{8} \cdot \sqrt{-1} = 2\sqrt{2} \cdot i = 2i\sqrt{2}. \][/tex]
Thus, [tex]\(\sqrt{-8} = 2i\sqrt{2}\)[/tex] is indeed correct.
### D. [tex]\(\sqrt{-8} = -2\sqrt{2}\)[/tex]
This statement is false. As seen above, [tex]\(\sqrt{-8} = 2i\sqrt{2}\)[/tex]. Since the square root of a negative number involves an imaginary unit [tex]\(i\)[/tex], [tex]\(\sqrt{-8}\)[/tex] cannot be a real number like [tex]\(-2\sqrt{2}\)[/tex].
### E. [tex]\(\sqrt{-16} = 4i\)[/tex]
This statement is true. Let's simplify [tex]\(\sqrt{-16}\)[/tex]:
[tex]\[ \sqrt{-16} = \sqrt{16 \cdot -1} = \sqrt{16} \cdot \sqrt{-1} = 4 \cdot i = 4i. \][/tex]
Thus, [tex]\(\sqrt{-16} = 4i\)[/tex] is correct.
### F. [tex]\(\sqrt{-16} = -4\)[/tex]
This statement is false. As established, [tex]\(\sqrt{-16} = 4i\)[/tex], which involves an imaginary unit. Therefore, [tex]\(\sqrt{-16}\)[/tex] cannot equal [tex]\(-4\)[/tex], a real number.
### Conclusion
The true statements are:
- C. [tex]\(\sqrt{-8} = 2i\sqrt{2}\)[/tex]
- E. [tex]\(\sqrt{-16} = 4i\)[/tex]
Therefore, the true statements are C and E.