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No, the student is incorrect. When solving the equation, the algebraic solution found is a = –2. However, this value must be checked against the restrictions of the original equation. Substituting –2 into the denominators (a + 2) and (a^2 – 4) results in division by zero, which is undefined. Therefore, a = –2 is an extraneous solution, and the original rational equation actually has no solution.
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Name the extraneous solution(s) to the equation.x = 0x = 1x = 8x = 0, x = 1, and x = 8x = 1 and x = 8There are no extraneous
How many extraneous solutions are there?zeroonetwo
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Name the true solution(s) to the equation.x = 0x = 1x = 8x = 0, x = 1, and x = 8✔ x = 1 and x = 8There are no solutions.
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