X + 1/(14 - X) = 1/(8X)
Understand the Problem
The question is asking for the solution to the equation X + 1/(14 - X) = 1/(8X). We will solve it by isolating the variable X and simplifying the equation step by step.
Answer
The solutions are $X = 1$, $X ≈ -0.052$, and $X ≈ -14.198$.
Answer for screen readers
The solutions to the equation are approximately $X = 1$, $X ≈ -0.052$, and $X ≈ -14.198$.
Steps to Solve
- Multiply through by common denominators
To eliminate the fractions, multiply both sides by the least common denominator, which is $8X(14 - X)$.
$$ (8X(14 - X)) \left( X + \frac{1}{14 - X} \right) = (8X(14 - X)) \left( \frac{1}{8X} \right) $$
This simplifies to:
$$ 8X^2(14 - X) + 8 = 14 - X $$
- Distribute and simplify the equation
Now, distribute on the left side:
$$ 8X^2(14) - 8X^3 + 8 = 14 - X $$
Rearranging terms gives us:
$$ -8X^3 + 8X^2 \cdot 14 + X - 14 + 8 = 0 $$
- Combine like terms
Combine like terms to simplify the polynomial:
$$ -8X^3 + 112X^2 + X - 6 = 0 $$
- Use synthetic division or numerical methods to factor the cubic
We can either use synthetic division, numerical methods, or graphing to solve.
Assuming one solution is $X = 1$, we divide:
$$ -8(1)^3 + 112(1)^2 + (1) - 6 = 0 $$
- Finding remaining roots
Having found one root, we can now use polynomial division or factorization to find other roots.
The equation can be factored as:
$$ -8(X - 1)(X^2 + 14.25X + 0.75) = 0 $$
- Solve for remaining roots using the quadratic formula
Now, apply the quadratic formula $X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 14.25$, and $c = 0.75$.
This will yield the other solutions for $X$.
The solutions to the equation are approximately $X = 1$, $X ≈ -0.052$, and $X ≈ -14.198$.
More Information
The primary solution can often give us insights into the behavior of the original equation, and other solutions may represent more complex interactions in the equation.
Tips
- Forgetting to distribute correctly after multiplying by the common denominator.
- Failing to combine like terms accurately, which can lead to incorrect polynomial formation.
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