Solve x2=36 by inspection. There are two real solutions. Enter the lesser number first.

Answer 1

To solve the equation \( x^2 = 36 \) by inspection, we look for numbers whose square equals 36.

The two numbers that satisfy this condition are:

• \( x = 6 \) (since \( 6^2 = 36 \))
• \( x = -6 \) (since \( (-6)^2 = 36 \))

Thus, the two real solutions are \( -6 \) and \( 6 \). The lesser number is:

\(-6\)

So, the solutions are \( -6 \) and \( 6 \). The final answer should be written as:

\(-6, 6\)

Answer 2

Solve x2=1625 by inspection. There are two real solutions. Enter the lesser number first. Leave the answers in simplest fraction form.

Answer 3

To solve the equation \( x^2 = 1625 \) by inspection, we need to find the numbers whose square equals 1625.

Firstly, we can calculate the square root of 1625:

\[ x = \pm \sqrt{1625} \]

To simplify \(\sqrt{1625}\), we will factor it:

\[ 1625 = 25 \times 65 \]

Since \( 25 = 5^2 \), we can rewrite it:

\[ \sqrt{1625} = \sqrt{25 \times 65} = \sqrt{25} \times \sqrt{65} = 5\sqrt{65} \]

Thus, we find the solutions to be:

\[ x = 5\sqrt{65} \quad \text{and} \quad x = -5\sqrt{65} \]

The lesser solution is:

\(-5\sqrt{65}\)

So, the two real solutions in simplest fraction form are:

\(-5\sqrt{65}, 5\sqrt{65}\)

Answer 4

Solve the following quadratic equation using square roots: (x+3)^2=36 .

Answer 5

To solve the equation \((x + 3)^2 = 36\) using square roots, follow these steps:

1. Take the square root of both sides: \[ x + 3 = \pm \sqrt{36} \]

2. Simplify the right side: \[ x + 3 = \pm 6 \]

3. Now, we will consider both cases:

   • Case 1: \(x + 3 = 6\) \[ x = 6 - 3 = 3 \]
   • Case 2: \(x + 3 = -6\) \[ x = -6 - 3 = -9 \]

4. Therefore, the solutions to the equation are: \[ x = 3 \quad \text{and} \quad x = -9 \]

The final answer is \( x = 3 \) and \( x = -9 \).