Answered

What is the solution for the equation $\frac{5}{3 b^3-2 b^2-5}=\frac{2}{b^3-2}$ ?


A$b=-4$ and $b=0$
B$b=-4$
C$b=0$ and $b=4$
D$b=4$

Verified Answer :

  • Cross-multiply the equation \(\frac{5}{3 b^3-2 b^2-5} = \frac{2}{b^3-2}\) to get \(5(b^3 - 2) = 2(3b^3 - 2b^2 - 5)\).
  • Expand and simplify to obtain the polynomial \(b^3 - 4b^2 = 0\).
  • Factor the expression as \(b^2(b - 4) = 0\) to find potential roots \(b = 0\) and \(b = 4\).
  • Verify that both solutions result in non-zero denominators, confirming the set of solutions is \(\boxed{b = 0 \text{ and } b = 4}\).

Explanation

  1. Analysis and Domain Restriction

We are given the rational equation:
\[\frac{5}{3b^3 - 2b^2 - 5} = \frac{2}{b^3 - 2}\]
Our goal is to find all values of \(b\) that satisfy this equation. First, we identify the domain of the equation by ensuring the denominators are non-zero:

  1. \(3b^3 - 2b^2 - 5 eq 0\)
  2. \(b^3 - 2 eq 0\)

We will check any potential solutions against these conditions at the end.

  1. Simplifying the Equation

To solve the rational equation, we cross-multiply the two sides to eliminate the denominators:
\[5(b^3 - 2) = 2(3b^3 - 2b^2 - 5)\]
Expanding both sides gives:
\[5b^3 - 10 = 6b^3 - 4b^2 - 10\]
Next, we rearrange the equation by subtracting \(5b^3 - 10\) from both sides to group the terms on one side:
\[0 = (6b^3 - 5b^3) - 4b^2 - 10 + 10\]
\[0 = b^3 - 4b^2\]
This simplifies to the polynomial equation:
\[b^3 - 4b^2 = 0\]

  1. Solving and Verifying Solutions

We factor the polynomial \(b^3 - 4b^2 = 0\) by pulling out the greatest common factor, \(b^2\):
\[b^2(b - 4) = 0\]
Setting each factor equal to zero provides the potential solutions:

  1. \(b^2 = 0 \implies b = 0\)
  2. \(b - 4 = 0 \implies b = 4\)

Now, we must verify these solutions against the original denominators:
For \(b = 0\):
Denominator 1: \(3(0)^3 - 2(0)^2 - 5 = -5 eq 0\)
Denominator 2: \(0^3 - 2 = -2 eq 0\)
For \(b = 4\):
Denominator 1: \(3(4)^3 - 2(4)^2 - 5 = 3(64) - 32 - 5 = 192 - 37 = 155 eq 0\)
Denominator 2: \(4^3 - 2 = 64 - 2 = 62 eq 0\)
Both values are valid solutions.

Examples

In engineering, rational equations often arise when calculating the ratio of two different physical quantities, such as the efficiency of two power sources or the distribution of current in a circuit. Solving these equations helps determine the specific conditions or parameters (like the value of \(b\)) where these two systems reach an equilibrium, represented by an equal ratio of their properties. Similar to this problem, ensuring the system operates within valid boundaries is crucial for real-world functionality.

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