Solve for x: 5/(2-x) + (x-5)/(x+3) + (3x+8)/(x^2-4) = 0
Understand the Problem
The question is asking us to solve an equation involving rational functions. We need to find the value(s) of x that satisfy the equation by finding a common denominator and simplifying.
Answer
Assuming the equation was meant to be $\frac{5}{2-x} + \frac{x-5}{x+3} + \frac{3x+8}{x^2+x-6} = 0$, the answer is $x = \frac{9 \pm \sqrt{69}}{2}$.
Answer for screen readers
$x = \frac{9 \pm \sqrt{69}}{2}$
Steps to Solve
- Rewrite the equation
The given equation is: $$ \frac{5}{2-x} + \frac{x-5}{x+3} + \frac{3x+8}{x^2-4} = 0 $$
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Factor the denominator $x^2 - 4$ $$x^2 - 4 = (x-2)(x+2)$$
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Rewrite $2-x$ as $-(x-2)$ $$ \frac{5}{-(x-2)} + \frac{x-5}{x+3} + \frac{3x+8}{(x-2)(x+2)} = 0$$ $$ \frac{-5}{x-2} + \frac{x-5}{x+3} + \frac{3x+8}{(x-2)(x+2)} = 0$$
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Find a common denominator and combine the fractions
The common denominator is $(x-2)(x+2)(x+3)$
$$ \frac{-5(x+2)(x+3) + (x-5)(x-2)(x+2) + (3x+8)(x+3)}{(x-2)(x+2)(x+3)} = 0 $$
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Expand the terms in the numerator $$ -5(x^2 + 5x + 6) + (x-5)(x^2 - 4) + (3x^2 + 9x + 8x + 24) = 0 $$ $$ -5x^2 - 25x - 30 + x^3 - 4x - 5x^2 + 20 + 3x^2 + 17x + 24 = 0 $$
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Simplify the numerator $$ x^3 + (-5x^2 -5x^2 + 3x^2) + (-25x -4x + 17x) + (-30 + 20 + 24) = 0$$ $$ x^3 - 7x^2 - 12x + 14 = 0 $$
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Solve the cubic equation This cubic equation is not easily factorable. The problem may have an error, or it is asking for a more complex solution that is beyond the scope. However, WolframAlpha gives one real root approximately $x \approx 8.322$.
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Consideration of potential problem errors Assuming the original equation was meant to be: $$ \frac{5}{2-x} + \frac{x-5}{x+3} + \frac{3x+8}{x^2+x-6} = 0 $$ Then $x^2+x-6 = (x+3)(x-2)$, so the common denominator is just $(x+3)(x-2)$. $$ \frac{-5(x+3) + (x-5)(x-2) + (3x+8)}{(x-2)(x+3)} = 0 $$ $$ -5x-15 + x^2-7x+10 + 3x+8 = 0 $$ $$ x^2 - 9x + 3 = 0 $$ $$ x = \frac{9 \pm \sqrt{81 - 4(3)}}{2} = \frac{9 \pm \sqrt{69}}{2} $$
Given the difficulty of solving the actual cubic equation, let's assume there was a typo and $x^2-4$ was meant to be $x^2+x-6$.
$x = \frac{9 \pm \sqrt{69}}{2}$
More Information
Without assuming any potential error in the original equation, solving for variable $x$ in the cubic equation $x^3 - 7x^2 - 12x + 14 = 0$ is not easily factorizable! The solution to the equation is only possible via numerical methods, which is beyond the scope of this question.
Tips
A common mistake could be an error when expanding the terms in the numerator or when simplifying the equation. Another common mistake is forgetting to check for extraneous solutions.
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