Answer :
To determine which solution to the equation [tex]\(\frac{3}{2g+8} = \frac{g+2}{g^2-16}\)[/tex] is extraneous, we need to examine the solutions carefully. Let’s go through the steps:
1. Factorize the denominator on the right side:
[tex]\[ g^2 - 16 = (g - 4)(g + 4) \][/tex]
2. Identify possible solutions:
When solving the equation, we solve for [tex]\(g\)[/tex], resulting in potential solutions [tex]\(g = -4\)[/tex] and [tex]\(g = 16\)[/tex].
3. Determine if any solutions lead to division by zero:
- For [tex]\(g = -4\)[/tex]:
[tex]\[ 2g + 8 = 2(-4) + 8 = -8 + 8 = 0 \][/tex]
Here, the denominator [tex]\(2g + 8\)[/tex] becomes zero, which makes the fraction [tex]\(\frac{3}{2g+8}\)[/tex] undefined. Hence, [tex]\(g = -4\)[/tex] is an extraneous solution.
- For [tex]\(g = 16\)[/tex]:
[tex]\[ g^2 - 16 = 16^2 - 16 = 256 - 16 = 240 \neq 0 \][/tex]
And also:
[tex]\[ 2g + 8 = 2(16) + 8 = 32 + 8 = 40 \][/tex]
Neither denominator becomes zero for [tex]\(g = 16\)[/tex], so this is not an extraneous solution.
4. Conclusion:
From the above analysis, we see that substituting [tex]\(g = -4\)[/tex] leads to a denominator of zero and is therefore an extraneous solution.
Thus, the extraneous solution to the equation [tex]\(\frac{3}{2g+8} = \frac{g+2}{g^2-16}\)[/tex] is:
[tex]\[ g = -4 \][/tex]
1. Factorize the denominator on the right side:
[tex]\[ g^2 - 16 = (g - 4)(g + 4) \][/tex]
2. Identify possible solutions:
When solving the equation, we solve for [tex]\(g\)[/tex], resulting in potential solutions [tex]\(g = -4\)[/tex] and [tex]\(g = 16\)[/tex].
3. Determine if any solutions lead to division by zero:
- For [tex]\(g = -4\)[/tex]:
[tex]\[ 2g + 8 = 2(-4) + 8 = -8 + 8 = 0 \][/tex]
Here, the denominator [tex]\(2g + 8\)[/tex] becomes zero, which makes the fraction [tex]\(\frac{3}{2g+8}\)[/tex] undefined. Hence, [tex]\(g = -4\)[/tex] is an extraneous solution.
- For [tex]\(g = 16\)[/tex]:
[tex]\[ g^2 - 16 = 16^2 - 16 = 256 - 16 = 240 \neq 0 \][/tex]
And also:
[tex]\[ 2g + 8 = 2(16) + 8 = 32 + 8 = 40 \][/tex]
Neither denominator becomes zero for [tex]\(g = 16\)[/tex], so this is not an extraneous solution.
4. Conclusion:
From the above analysis, we see that substituting [tex]\(g = -4\)[/tex] leads to a denominator of zero and is therefore an extraneous solution.
Thus, the extraneous solution to the equation [tex]\(\frac{3}{2g+8} = \frac{g+2}{g^2-16}\)[/tex] is:
[tex]\[ g = -4 \][/tex]
