Pergunta

Solve for all values of x. 1+(7)/(x-4)=(2)/(x^2)-16 Answer Attemptiout of 2 Additional Solution (C) No Solution x=square
Solução

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AlbertoProfissional · Tutor por 6 anos
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To solve the equation \(1 + \frac{7}{x-4} = \frac{2}{x^2 - 16}\), we need to find the values of \(x\) that satisfy this equation.<br /><br />First, let's rewrite the equation in a more manageable form. Notice that \(x^2 - 16\) can be factored as \((x-4)(x+4)\). So, the equation becomes:<br /><br />\[1 + \frac{7}{x-4} = \frac{2}{(x-4)(x+4)}\]<br /><br />Next, let's multiply both sides of the equation by \((x-4)(x+4)\) to eliminate the denominators:<br /><br />\[(x-4)(x+4) \left(1 + \frac{7}{x-4}\right) = \frac{2}{(x-4)(x+4)} \cdot (x-4)(x+4)\]<br /><br />Simplifying both sides, we get:<br /><br />\[(x-4)(x+4) + 7(x+4) = 2\]<br /><br />Expanding and combining like terms, we have:<br /><br />\[x^2 - 16 + 7x + 28 = 2\]<br /><br />\[x^2 + 7x + 12 = 0\]<br /><br />Now, we can factor the quadratic equation:<br /><br />\[(x + 3)(x + 4) = 0\]<br /><br />Setting each factor equal to zero gives us the solutions:<br /><br />\[x + 3 = 0 \quad \text{or} \quad x + 4 = 0\]<br /><br />\[x = -3 \quad \text{or} \quad x = -4\]<br /><br />However, we need to check these solutions in the original equation to ensure they are valid. Substituting \(x = -3\) into the original equation, we find that it does not satisfy the equation. Therefore, the only valid solution is \(x = -4\).<br /><br />So, the solution to the equation is \(x = -4\).
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