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Solve for all values of
x.

(x-5)/(x+3)=(-1)/(x)
Answer:
x=

Solve for all values of $x$ . $\newline$ $\frac{x-5}{x+3}=\frac{-1}{x}$ $\newline$ Answer: $x=$

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Find the roots of factored polynomials

Full solution

Q. Solve for all values of

x

.

\newline

\frac{x-5}{x+3}=\frac{-1}{x}

\newline

Answer:

x=

Identify Equation: First, we need to identify the equation we are solving: $(x-5)/(x+3)=(-1)/(x)$ . We will solve for $x$ by finding a common denominator and then cross-multiplying to eliminate the fractions.
Find Common Denominator: The common denominator between $(x+3)$ and $x$ is $x(x+3)$ . We will multiply both sides of the equation by this common denominator to clear the fractions.
Clear Fractions: Multiplying both sides of the equation by $x(x+3)$ , we get: $\newline$ $(\frac{x-5}{x+3}) \cdot x(x+3) = (\frac{-1}{x}) \cdot x(x+3)$ $\newline$ This simplifies to: $\newline$ $x(x-5) = -1(x+3)$
Distribute and Simplify: Now we distribute on both sides of the equation: $x^2 - 5x = -x - 3$
Combine Like Terms: Next, we bring all terms to one side of the equation to set it equal to zero: $x^2 - 5x + x + 3 = 0$
Factor Quadratic Equation: Combine like terms: $x^2 - 4x + 3 = 0$
Solve for x: Now we factor the quadratic equation: $\newline$ $(x - 3)(x - 1) = 0$
Check Solutions: Set each factor equal to zero and solve for $x$ : $x - 3 = 0$ or $x - 1 = 0$
Check Solutions: Set each factor equal to zero and solve for $x$ : $x - 3 = 0$ or $x - 1 = 0$ Solving each equation gives us the values of $x$ : $x = 3$ or $x = 1$
Check Solutions: Set each factor equal to zero and solve for $x$ : $x - 3 = 0$ or $x - 1 = 0$ Solving each equation gives us the values of $x$ : $x = 3$ or $x = 1$ However, we must check these solutions against the original equation to ensure they do not make any denominator zero. The original equation has denominators $(x+3)$ and $x$ , so we must exclude any solutions where $x = -3$ or $x = 0$ .
Check Solutions: Set each factor equal to zero and solve for $x$ : $x - 3 = 0$ or $x - 1 = 0$ Solving each equation gives us the values of $x$ : $x = 3$ or $x = 1$ However, we must check these solutions against the original equation to ensure they do not make any denominator zero. The original equation has denominators $(x+3)$ and $x$ , so we must exclude any solutions where $x = -3$ or $x = 0$ .Checking the solutions, we see that neither $x = 3$ nor $x = 1$ makes any denominator zero, so both are valid solutions.

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