What are the solutions, if any, of the equation [tex]$x^2 + 38 = 16$[/tex]?

A. [tex]$x = \pm \sqrt{22}$[/tex]
B. [tex][tex]$x = \pm 3 \sqrt{6}$[/tex][/tex]
C. [tex]$x = 4 \pm \sqrt{38}$[/tex]
D. This equation has no real solutions.



Answer :

To solve the equation [tex]\( x^2 + 38 = 16 \)[/tex], let's go through the solution step-by-step:

1. Isolate the [tex]\( x^2 \)[/tex] term:
[tex]\[ x^2 + 38 = 16 \][/tex]
First, subtract 38 from both sides of the equation:
[tex]\[ x^2 + 38 - 38 = 16 - 38 \][/tex]
Simplifying this gives:
[tex]\[ x^2 = -22 \][/tex]

2. Analyze the equation [tex]\( x^2 = -22 \)[/tex]:
In this step, observe that [tex]\( x^2 = -22 \)[/tex] implies that the square of [tex]\( x \)[/tex] equals a negative number.

In the field of real numbers, there is no real number that, when squared, results in a negative value. Therefore, there are no real solutions to this equation.

3. Conclusion for real numbers:
Because it's impossible for [tex]\( x^2 \)[/tex] to equal a negative number in the real number system, we conclude that this equation has no real solutions.

Thus, the final answer is:
[tex]\[ \text{This equation has no real solutions.} \][/tex]