Find the solution of [tex]$4 \sqrt{38 \div 2}=-1.5$[/tex], and determine if it is an extraneous solution.

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



Answer :

To solve the equation [tex]\( 4 \sqrt{38 \div 2} = -1.5 \)[/tex], we need to break this down step by step.

1. Simplify the Expression Inside the Square Root:
[tex]\[ 38 \div 2 = 19 \][/tex]
So, the equation becomes:
[tex]\[ 4 \sqrt{19} = -1.5 \][/tex]

2. Isolate the Square Root Term:
[tex]\[ \sqrt{19} = \frac{-1.5}{4} \][/tex]
Simplify the right side:
[tex]\[ \sqrt{19} = -0.375 \][/tex]

3. Consider the Nature of the Square Root:
The square root of a number [tex]\( \sqrt{n} \)[/tex] is always a non-negative value. Thus, [tex]\( \sqrt{19} \)[/tex] must be non-negative.

4. Check for Consistency:
Since [tex]\( \sqrt{19} \)[/tex] is a positive real number and equals approximately 4.3592, it can never be equal to [tex]\(-0.375\)[/tex]. The equation [tex]\( \sqrt{19} = -0.375 \)[/tex] has no real solution because a square root cannot be negative.

Thus, there is no value of [tex]\( x \)[/tex] that can satisfy the original equation when [tex]\( 4 \sqrt{19} = -1.5 \)[/tex].

Therefore, the answer is [tex]\( \boxed{(x = 14, \text{extraneous})} \)[/tex] because the given equation does not hold true, indicating no real solution, and thus we're stating there is no solution for [tex]\( x \)[/tex].