Consider the following equation.

Question

x^2 = 2

The equation above has_____ (A: two B: one C: three D: no) solution(s).
When x = , x is_____(A: Not a B: a) solution to the equation.
The solution(s) for the given equation are____ (A: rational B: irrational) number(s).

Answer 1

Let's analyze the equation $x^2=2$.

First, we can determine the number of solutions:

1. Taking the square root of both sides, we get: $$x=\sqrt{2}\text{or}x=-\sqrt{2}$$ Thus, there are two solutions: $\sqrt{2}$ and $-\sqrt{2}$.

So, the equation above has (A: two) solution(s).

Next, when $x=\sqrt{2}$ or $x=-\sqrt{2}$, we check if it is a solution:

• Since both values satisfy the original equation $x^2=2$, $x=\sqrt{2}$ (and $x=-\sqrt{2}$) is (B: a) solution to the equation.

Finally, we need to determine the nature of the solutions:

• Both $\sqrt{2}$ and $-\sqrt{2}$ are not rational numbers (they cannot be expressed as a fraction of two integers). Therefore, the solutions for the given equation are (B: irrational) number(s).

Putting this all together:

1. The equation above has (A: two) solution(s).
2. When $x=\sqrt{2}$, $x$ is (B: a) solution to the equation.
3. The solution(s) for the given equation are (B: irrational) number(s).

Answer 2

Simply square root of 16

Answer 3

The square root of 16 is $4$.

This is because $4\times 4=16$.

So, $\sqrt{16}=4$.