Which of the following equations has both -5 and 5 as possible values of [tex]$z$[/tex]?

Choose all answers that apply:

A. [tex]$z^2 = 25$[/tex]

B. [tex][tex]$z^3 = 125$[/tex][/tex]

C. None of the above



Answer :

Let's analyze each equation to determine which values of [tex]\( z \)[/tex] are valid solutions.

### Equation (A): [tex]\( z^2 = 36 \)[/tex]

First, solve the equation:
[tex]\[ z^2 = 36 \][/tex]

Take the square root of both sides of the equation:
[tex]\[ z = \pm \sqrt{36} \][/tex]
[tex]\[ z = \pm 6 \][/tex]

Thus, the solutions to [tex]\( z^2 = 36 \)[/tex] are [tex]\( z = 6 \)[/tex] and [tex]\( z = -6 \)[/tex]. Therefore, neither [tex]\( z = -5 \)[/tex] nor [tex]\( z = 5 \)[/tex] are solutions to this equation.

### Equation (B): [tex]\( z^3 = 125 \)[/tex]

Next, solve the equation:
[tex]\[ z^3 = 125 \][/tex]

Take the cube root of both sides of the equation:
[tex]\[ z = \sqrt[3]{125} \][/tex]
[tex]\[ z = 5 \][/tex]

Thus, the only solution to [tex]\( z^3 = 125 \)[/tex] is [tex]\( z = 5 \)[/tex], and not [tex]\( z = -5 \)[/tex].

### None of the above

Reviewing the solutions to the equations:
- For [tex]\( z^2 = 36 \)[/tex], the valid solutions are [tex]\( z = 6 \)[/tex] and [tex]\( z = -6 \)[/tex].
- For [tex]\( z^3 = 125 \)[/tex], the valid solution is [tex]\( z = 5 \)[/tex].

Neither equation (A) nor equation (B) allows both [tex]\( z = -5 \)[/tex] and [tex]\( z = 5 \)[/tex] as solutions.

Therefore, the correct answer is:

[tex]\[ \boxed{\text{None of the above}} \][/tex]

Hence, the appropriate choice is:
- (C) None of the above