Let's analyze the equation and determine if the student's claim is valid.
The equation given is:
[tex]\[ 5x^2 = 20 \][/tex]
To solve for [tex]\( x \)[/tex], we need to isolate [tex]\( x \)[/tex]. Follow these steps:
1. Divide both sides by 5 to simplify the equation:
[tex]\[ x^2 = \frac{20}{5} \][/tex]
[tex]\[ x^2 = 4 \][/tex]
2. Take the square root of both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \pm \sqrt{4} \][/tex]
The square root of 4 can be either positive or negative, therefore:
[tex]\[ x = 2 \quad \text{or} \quad x = -2 \][/tex]
Thus, the solutions to the equation [tex]\( 5x^2 = 20 \)[/tex] are [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex].
3. Evaluate the student's claim:
The student claims that [tex]\( x \)[/tex] must be equal to [tex]\( 2 \)[/tex]. While [tex]\( x = 2 \)[/tex] is indeed a solution to the equation, it is not the only solution. The full set of solutions includes [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex].
Therefore, the student's assertion that [tex]\( x \)[/tex] must be equal to [tex]\( 2 \)[/tex] alone is incomplete. The equation [tex]\( 5x^2 = 20 \)[/tex] has two valid solutions: [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex].
So, I would disagree with the student's statement because it does not take into account the negative solution [tex]\( x = -2 \)[/tex].
