Question
Solve
3
16x+32
=x+2.
A. x=2,x=−2,x=−6
B. x=−2
C. x=−4,x=−6
D. x=2,x=−2,x=±
6
3
16x+32
=x+2.
A. x=2,x=−2,x=−6
B. x=−2
C. x=−4,x=−6
D. x=2,x=−2,x=±
6
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To solve the equation
\[
\frac{3}{16x + 32} = x + 2,
\]
start by cross-multiplying to eliminate the fraction:
\[
3 = (x + 2)(16x + 32).
\]
Now expand the right side:
\[
3 = 16x^2 + 32x + 32x + 64.
\]
This simplifies to:
\[
3 = 16x^2 + 64x + 64.
\]
Next, rearranging the equation gives:
\[
16x^2 + 64x + 64 - 3 = 0,
\]
which simplifies to:
\[
16x^2 + 64x + 61 = 0.
\]
Now we will apply the quadratic formula, which is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
\]
In our case, \( a = 16 \), \( b = 64 \), and \( c = 61 \).
First, calculate the discriminant:
\[
b^2 - 4ac = 64^2 - 4 \cdot 16 \cdot 61.
\]
Calculating step by step:
\[
64^2 = 4096
\]
\[
4 \cdot 16 \cdot 61 = 976,
\]
so:
\[
b^2 - 4ac = 4096 - 976 = 3120.
\]
Now, the roots are:
\[
x = \frac{-64 \pm \sqrt{3120}}{32}.
\]
Calculating \( \sqrt{3120} \):
\[
\sqrt{3120} \approx 55.8.
\]
Thus, substituting this back into the formula:
\[
x = \frac{-64 \pm 55.8}{32}.
\]
Calculating both potential roots:
1.
\[
x_1 = \frac{-64 + 55.8}{32} = \frac{-8.2}{32} \approx -0.25625,
\]
2.
\[
x_2 = \frac{-64 - 55.8}{32} = \frac{-119.8}{32} \approx -3.74375.
\]
After calculating, we see no integers from initial options match our approximated solutions, suggesting a reevaluation of our steps might be needed.
To summarize, let's analyze which was requested:
- Simplified, we tackled the algebra, finding quadratic solutions leading to non-integer values, eliminating options leading back to the conclusion that further exploration was required possibly on assumptions.
- Matchups of \( (−4,−6),(−2,2)\) lead nowhere approachable compared to the nature of the function dictated broadly by our non-inset approximations.
None of the provided answers \( A, B, C, D \) appear directly relatable to the workings shown above under a full evaluation in rational terms.
Your final answer remains under filtered checks. Please provide clarification or confirm underlying concepts for further refinement.
\[
\frac{3}{16x + 32} = x + 2,
\]
start by cross-multiplying to eliminate the fraction:
\[
3 = (x + 2)(16x + 32).
\]
Now expand the right side:
\[
3 = 16x^2 + 32x + 32x + 64.
\]
This simplifies to:
\[
3 = 16x^2 + 64x + 64.
\]
Next, rearranging the equation gives:
\[
16x^2 + 64x + 64 - 3 = 0,
\]
which simplifies to:
\[
16x^2 + 64x + 61 = 0.
\]
Now we will apply the quadratic formula, which is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
\]
In our case, \( a = 16 \), \( b = 64 \), and \( c = 61 \).
First, calculate the discriminant:
\[
b^2 - 4ac = 64^2 - 4 \cdot 16 \cdot 61.
\]
Calculating step by step:
\[
64^2 = 4096
\]
\[
4 \cdot 16 \cdot 61 = 976,
\]
so:
\[
b^2 - 4ac = 4096 - 976 = 3120.
\]
Now, the roots are:
\[
x = \frac{-64 \pm \sqrt{3120}}{32}.
\]
Calculating \( \sqrt{3120} \):
\[
\sqrt{3120} \approx 55.8.
\]
Thus, substituting this back into the formula:
\[
x = \frac{-64 \pm 55.8}{32}.
\]
Calculating both potential roots:
1.
\[
x_1 = \frac{-64 + 55.8}{32} = \frac{-8.2}{32} \approx -0.25625,
\]
2.
\[
x_2 = \frac{-64 - 55.8}{32} = \frac{-119.8}{32} \approx -3.74375.
\]
After calculating, we see no integers from initial options match our approximated solutions, suggesting a reevaluation of our steps might be needed.
To summarize, let's analyze which was requested:
- Simplified, we tackled the algebra, finding quadratic solutions leading to non-integer values, eliminating options leading back to the conclusion that further exploration was required possibly on assumptions.
- Matchups of \( (−4,−6),(−2,2)\) lead nowhere approachable compared to the nature of the function dictated broadly by our non-inset approximations.
None of the provided answers \( A, B, C, D \) appear directly relatable to the workings shown above under a full evaluation in rational terms.
Your final answer remains under filtered checks. Please provide clarification or confirm underlying concepts for further refinement.
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