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Find the solution of [tex]$4 \sqrt{x+2}=-16$[/tex], and determine if it is an extraneous solution.

A. [tex]$x=14$[/tex]; extraneous
B. [tex]$x=14$[/tex]; not extraneous
C. [tex]$x=2$[/tex]; extraneous
D. [tex]$x=2$[/tex]; not extraneous


Sagot :

To solve the equation [tex]\(4 \sqrt{x + 2} = -16\)[/tex], follow these steps:

1. Isolate the square root term: The equation is already isolated with the square root term on one side.
[tex]\[ 4 \sqrt{x + 2} = -16 \][/tex]

2. Divide both sides by 4: Simplify the equation by dividing both sides by 4.
[tex]\[ \sqrt{x + 2} = -4 \][/tex]

3. Consider the nature of a square root: Remember that the square root of any real number is always non-negative. Therefore, the equation [tex]\(\sqrt{x + 2} = -4\)[/tex] has no solution because [tex]\(\sqrt{x + 2}\)[/tex] can never equal a negative number.

Therefore, the equation [tex]\(4 \sqrt{x + 2} = -16\)[/tex] has no solution. Since there are no solutions directly derived from the equation, there are no possibilities to be extraneous or valid.

Thus, the conclusion is:
- There are no valid solutions.
- The correct option is [tex]\(x =\)[/tex] None; not solvable.

According to the information provided above, none of the given options correctly represent the solution for the equation.
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