The statement is:
* (S) if $x^2 = 36$ then $x=6$

That's logically equivalent to:
* (S) $x^2 \ne 36 \lor x=6$

Before proceeding, I should say that I understand what the question is *trying* to test the student on. I don't think it's obvious which option is correct. We need to be clear about two things: domain and quantification.

## Domain

One of the answers mentions "integer", so we could consider two cases: $x \in \mathbb{Z}$ and $x \in \mathbb{N}$. Each of these will result in different truth values for S. But we don't need the full sets of integers or natural numbers, so let's focus on these:
* $D_1 = \{-6, +6\}$
* $D_2 = \{+6\}$
* $D_3 = \{-6\}$

## Quantification

There's an implicit quantifier around S and the answer depends on which kind it is:
* ($S_1$) $\exists_{x\in D}\big[x^2 \ne 36 \lor x=6\big]$
* ($S_2$) $\forall_{x\in D}\big[x^2 \ne 36 \lor x=6\big]$

## Answer

### A. Suppose the statement is $S_1$

The statement is:
* ($S_1$) $\exists_{x\in D}\big[x^2 \ne 36 \lor x=6\big]$

##### 1. if $D_1 = \{-6, +6\}$

1. $6 \in D_1$
2. $\exists_{x \in D_1}[x = 6]$
3. $\exists_{x\in D_1}\big[x^2 \ne 36 \lor x=6\big]$ (disjunction-introduction)
4. $S_1$ is **true**

##### 2. if $D_2 = \{+6\}$

1. $6 \in D_2$
2. $\exists_{x \in D_2}[x = 6]$
3. $\exists_{x\in D_2}\big[x^2 \ne 36 \lor x=6\big]$ (disjunction-introduction)
4. $S_1$ is **true**

Exactly the same argument as in (A.1).

##### 3. if $D_3 = \{-6\}$

There is only one object here, namely $-6$ and it doesn't satisfy either of these:
1. $x = 6$
2. $x^2 \ne 36$

It doesn't satisfy (1) because $-6 \ne 6$ and (2) because $(-6)^2$ is actually 36. Therefore $S_1$ is **false**

##### 4. if $\mathbb{N}$

Since $6 \in \mathbb{N}$, $S_1$ is **true**.

##### 5. if $\mathbb{Z}$

Since $6 \in \mathbb{Z}$, $S_1$ is **true**.

### B. Suppose the statement is $S_2$

* ($S_2$) $\forall_{x\in D}\big[x^2 \ne 36 \lor x=6\big]$

##### 1. if $D_1 = \{-6, +6\}$

1. pick $x = -6$
2. $x \ne 6$ (since $-6 \ne 6$)
3. $x^2 = 36$
4. $x^2 = 36 \land x\ne6$ (combining 2 and 3)
5. $S_2$ is **false**

##### 2. if $D_2 = \{+6\}$

+6 is the only member of this domain, so the formula is satisfied for all $x \in D_2$. $S_2$ is **true**.

##### 3. if $D_3 = \{-6\}$

$S_2$ is **false** (same argument as in B.1)

##### 4. if $\mathbb{N}$

This is an interesting case:

* pick arbitrary $a \in \mathbb{N}$
* either $a = 6 \lor a\ne 6$
* case $a=6$
* $S_2$ is **true**
* case $a\ne6$
* suppose, for contradiction, that $x^2 = 36$
* instantiating that we get: $a^2 = 36$
* therefore $a = \sqrt{36}$
* square root is a *function* on $\mathbb{N}$, so $\sqrt{36} = 6$
* so $a = 6$
* contradiction between $a=6$ and $a\ne 6$
* therefore, $x^2 \ne 36$
* therefore, $S_2$ is **true**
* $S_2$ is **true** (by case analysis)

##### 5. if $\mathbb{Z}$

$S_2$ is **false** (same argument as in B.1)

# Summary

As I explained at the beginning of this note, the answer depends on the domain and the quantification. The possibilities with their answers are summarized below:

| $D:$| $\{-6, 6\}$ | $\{6\}$ | $\{-6\}$ | $\mathbb{N}$ | $\mathbb{Z}$
| - | - | - | - | - | -
| $S_1=\exists_{x\in D}\big[x^2 \ne 36 \lor x=6\big]$ | T|T|F |T |T
| $S_2=\forall_{x\in D}\big[x^2 \ne 36 \lor x=6\big]$ | F|T |F |T |F

I think the *intended* question and answer were:
* quantification: $S_2$
* domain: $D=\mathbb{Z}$ (but any set with $-6$ in it would suffice)
* answer: false

But as you can see, the answer depends on things. The lesson of course is that squaring is not invertible on $\mathbb{Z}$, that the square root "function" is actually a many-valued *relation* (if we think of its codomain as a cartesian product though, it does behave like a function taking numbers (e.g. $36$) as input and giving pairs (e.g. $(-6, +6)$ as output).