After multiplying each side of the equation by the LCD and simplifying, the resulting equation is 3x² – 20x + 12 = 03x² – 20x – 12 = 0✔ 3x² + 20x + 12 = 03x² + 20x – 12 = 0.
1 x + 3
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x = 1 is a solution.x = 0 is a solution.x = –1 is a solution.x = –6 is a solution.
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No, the student is incorrect. While solving the equation algebraically results in the value -2, this value is an extraneous solution. If you substitute -2 back into the original equation, the denominators (a + 2) and (a^2 - 4) both become zero. Since division by zero is undefined, -2 cannot be a solution, and the equation actually has no solution.
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What are the solutions to the equation?



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-4
Name the true solution(s) to the equation.X x = 0x = 1x = 8x = 0, x = 1, and x = 8✔ x = 1 and x = 8There are no solutions.
Name the extraneous solution(s) to the equation.✔ x = 0x = 1x = 8x = 0, x = 1, and x = 8x = 1 and x = 8X There are no extraneous
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