Solve the following equations by inspection. There are two real solutions. Enter the lesser number first.

Question

(1 point)
1) x2=36
The solutions are:
and

2) x2=81
The solutions are:
and

3) x2=4
The solutions are:
and

4) x2=49
The solutions are:
and

5) x2=25
The solutions are:
and

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Answer 1

To solve the equations \(x^2 = k\), where \(k\) is a constant, we can take the square root of both sides. However, since we are looking for both positive and negative solutions, the solutions can be expressed as:

\[ x = \pm \sqrt{36}, \pm \sqrt{81}, \pm \sqrt{4}, \pm \sqrt{49}, \pm \sqrt{25} \]

Now, let's find the solutions for each equation:

1. \(x^2 = 36\)

   • Solutions are: \(-6\) and \(6\)

2. \(x^2 = 81\)

   • Solutions are: \(-9\) and \(9\)

3. \(x^2 = 4\)

   • Solutions are: \(-2\) and \(2\)

4. \(x^2 = 49\)

   • Solutions are: \(-7\) and \(7\)

5. \(x^2 = 25\)

   • Solutions are: \(-5\) and \(5\)

Now, we can summarize the solutions:

1. The solutions are: \(-6\) and \(6\)
2. The solutions are: \(-9\) and \(9\)
3. The solutions are: \(-2\) and \(2\)
4. The solutions are: \(-7\) and \(7\)
5. The solutions are: \(-5\) and \(5\)