First simplify the equation using common denominators. Next, solve using the quadratic formula to find potential values for x. Then check these solutions in the original equation; any root that makes the denominator zero is an extraneous solution, as it does not satisfy the original equation.
To solve for extraneous solutions in the equation (3)/(x−1)=−(6)/(x²−1), we first need to multiply both sides by the common denominator to simplify the fractions: 3(x²-1) = -6(x-1). This simplifies to 3x²-3 = -6x+6. Adding like terms, we get 3x² + 6x - 3 = 0.
We can then use the quadratic formula x = [-b ± sqrt(b² - 4ac)] / (2a) to solve for x. Substituting values from the equation where a=3 b=6 and c=-3 we get the roots of the equation.
However, an extraneous solution is a root that doesn't satisfy the original, rational equation. So we must check the roots in the original equation to verify if they make any denominator zero, as division by zero is undefined in mathematics. Any root that does that is an extraneous solution.
Learn more about Extraneous Solutions