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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).

3 answers

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).
Simply square root of 16

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

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

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