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Sagot :
To determine the domain restrictions for the given rational equation:
[tex]\[ \frac{2}{3 x^2} + \frac{x}{2} = \frac{x+5}{x} \][/tex]
we need to identify the values of [tex]\( x \)[/tex] that make any denominator in the equation equal to zero. Denominators that are zero would make the entire equation undefined at those values of [tex]\( x \)[/tex].
Let's analyze the denominators in the equation:
1. The first term [tex]\(\frac{2}{3x^2}\)[/tex]:
- The denominator is [tex]\(3x^2\)[/tex].
- This will be zero when [tex]\(x = 0\)[/tex].
2. The second term [tex]\(\frac{x}{2}\)[/tex]:
- The denominator is [tex]\(2\)[/tex], which is never zero for any value of [tex]\( x \)[/tex]. Hence, no restriction from this term.
3. The third term [tex]\(\frac{x+5}{x}\)[/tex]:
- The denominator is [tex]\(x\)[/tex].
- This will be zero when [tex]\(x = 0\)[/tex].
Combining these observations, the only value of [tex]\( x \)[/tex] that would make any of the denominators zero is:
[tex]\[ x = 0 \][/tex]
Therefore, [tex]\( x = 0 \)[/tex] is a restriction on the domain. We should now look at our given answer choices and select all that apply:
A. There are no restrictions on the domain. - This is incorrect because we found that [tex]\( x = 0 \)[/tex] is a restriction.
B. [tex]\( x = -5 \)[/tex] - This is incorrect because [tex]\( x = -5 \)[/tex] does not make any denominator zero.
C. [tex]\( x = 2 \)[/tex] - This is incorrect because [tex]\( x = 2 \)[/tex] does not make any denominator zero.
D. [tex]\( x = 0 \)[/tex] - This is correct because [tex]\( x = 0 \)[/tex] does make one of the denominators zero.
Thus, the correct answer is:
D. [tex]\( x = 0 \)[/tex]
[tex]\[ \frac{2}{3 x^2} + \frac{x}{2} = \frac{x+5}{x} \][/tex]
we need to identify the values of [tex]\( x \)[/tex] that make any denominator in the equation equal to zero. Denominators that are zero would make the entire equation undefined at those values of [tex]\( x \)[/tex].
Let's analyze the denominators in the equation:
1. The first term [tex]\(\frac{2}{3x^2}\)[/tex]:
- The denominator is [tex]\(3x^2\)[/tex].
- This will be zero when [tex]\(x = 0\)[/tex].
2. The second term [tex]\(\frac{x}{2}\)[/tex]:
- The denominator is [tex]\(2\)[/tex], which is never zero for any value of [tex]\( x \)[/tex]. Hence, no restriction from this term.
3. The third term [tex]\(\frac{x+5}{x}\)[/tex]:
- The denominator is [tex]\(x\)[/tex].
- This will be zero when [tex]\(x = 0\)[/tex].
Combining these observations, the only value of [tex]\( x \)[/tex] that would make any of the denominators zero is:
[tex]\[ x = 0 \][/tex]
Therefore, [tex]\( x = 0 \)[/tex] is a restriction on the domain. We should now look at our given answer choices and select all that apply:
A. There are no restrictions on the domain. - This is incorrect because we found that [tex]\( x = 0 \)[/tex] is a restriction.
B. [tex]\( x = -5 \)[/tex] - This is incorrect because [tex]\( x = -5 \)[/tex] does not make any denominator zero.
C. [tex]\( x = 2 \)[/tex] - This is incorrect because [tex]\( x = 2 \)[/tex] does not make any denominator zero.
D. [tex]\( x = 0 \)[/tex] - This is correct because [tex]\( x = 0 \)[/tex] does make one of the denominators zero.
Thus, the correct answer is:
D. [tex]\( x = 0 \)[/tex]
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