Question

1. select all the values that represent solutions to the equation $x^2 = 13$.

$x = \frac{13}{2}$ (a) $x = \sqrt{13}$ (b)
$x = \sqrt{-13}$ (c) $x = -\sqrt{-13}$ (d)
$x = -\sqrt{13}$ (e) $x = 13^2$ (f)

Explanation:

Step1: Solve the equation \(x^2 = 13\)

To solve for \(x\), we take the square root of both sides. Remember that when we take the square root of a number, we get both a positive and a negative root. So, \(x=\pm\sqrt{13}\), which means \(x = \sqrt{13}\) or \(x=-\sqrt{13}\).

Step2: Analyze each option

• Option A: \(x = \frac{13}{2}\). Squaring this gives \((\frac{13}{2})^2=\frac{169}{4}

eq13\), so it's not a solution.

• Option B: \(x = \sqrt{13}\). Squaring \(\sqrt{13}\) gives \((\sqrt{13})^2 = 13\), so it is a solution.
• Option C: \(x=\sqrt{- 13}\). The square root of a negative number is not a real number, and our original equation \(x^2 = 13\) (where 13 is positive) has real solutions, so this is not a solution.
• Option D: \(x = -\sqrt{-13}\). Again, this involves the square root of a negative number, so it's not a real solution for our equation.
• Option E: \(x=-\sqrt{13}\). Squaring \(-\sqrt{13}\) gives \((-\sqrt{13})^2=(- 1)^2\times(\sqrt{13})^2 = 13\), so it is a solution.
• Option F: \(x = 13^2=169\). Squaring 169 gives \(169^2

eq13\), so it's not a solution.

Answer:

B. \(x = \sqrt{13}\), E. \(x=-\sqrt{13}\)