Pregunta
4) \( -3(4 b+1)^{\frac{3}{2}}+5=-1024 \)
Ask by Cross Bradley. in the United States
Feb 03,2025
Solución de tutoría real
Respuesta verificada por el tutor
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There are no real solutions for \( b \).
Solución
Solve the equation \( -3(4b+1)^{\frac{3}{2}}+5=-1024 \).
Solve the equation by following steps:
- step0: Solve for \(b\):
\(-3\left(4b+1\right)^{\frac{3}{2}}+5=-1024\)
- step1: Find the domain:
\(-3\left(4b+1\right)^{\frac{3}{2}}+5=-1024,b\geq -\frac{1}{4}\)
- step2: Move the expression to the left side:
\(-3\left(4b+1\right)^{\frac{3}{2}}+1029=0\)
- step3: Factor the expression:
\(3\left(7-4b-1\right)\left(49+7\left(4b+1\right)+\left(4b+1\right)^{2}\right)=0\)
- step4: Multiply both sides:
\(3\left(7-4b-1\right)\left(49+7\left(4b+1\right)+\left(4b+1\right)^{2}\right)\times \frac{1}{3}=0\times \frac{1}{3}\)
- step5: Calculate:
\(\left(7-4b-1\right)\left(49+7\left(4b+1\right)+\left(4b+1\right)^{2}\right)=0\)
- step6: Separate into possible cases:
\(\begin{align}&7-4b-1=0\\&49+7\left(4b+1\right)+\left(4b+1\right)^{2}=0\end{align}\)
- step7: Solve the equation:
\(\begin{align}&b=\frac{3}{2}\\&b \notin \mathbb{R}\end{align}\)
- step8: Find the union:
\(b=\frac{3}{2}\)
- step9: Check if the solution is in the defined range:
\(b=\frac{3}{2},b\geq -\frac{1}{4}\)
- step10: Find the intersection:
\(b=\frac{3}{2}\)
- step11: Check the solution:
\(b \in \varnothing \)
The solution to the equation \( -3(4b+1)^{\frac{3}{2}}+5=-1024 \) is that there are no real solutions for \( b \).
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The Deep Dive
To solve \( -3(4b + 1)^{\frac{3}{2}} + 5 = -1024 \), first, isolate the term with \( b \): 1. Subtract 5 from both sides: \[ -3(4b + 1)^{\frac{3}{2}} = -1029 \] 2. Divide by -3: \[ (4b + 1)^{\frac{3}{2}} = 343 \] 3. Take both sides to the power of \( \frac{2}{3} \): \[ 4b + 1 = 7 \] 4. Subtract 1 from both sides: \[ 4b = 6 \] 5. Finally, divide by 4: \[ b = \frac{3}{2} \] Thus, the solution is \( b = \frac{3}{2} \).
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