Question

Consider the following equation.

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

Simply square root of 16

Answer 2

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:
\[
x = \sqrt{2} \quad \text{or} \quad x = -\sqrt{2}
\]
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 3

The square root of 16 is \( 4 \).

This is because \( 4 \times 4 = 16 \).

So, \( \sqrt{16} = 4 \).