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Answer:

The student is not correct in claiming that 𝑎 = − 2 a=−2 is a solution. The original equation is 3 𝑎 + 2 − 6 𝑎 𝑎 2 − 4 = 1 𝑎 − 2 a 2 −4 3a+2−6a ​ = a−2 1 ​ . Simplifying the numerator on the left gives − 3 𝑎 + 2 −3a+2, and factoring the denominator gives ( 𝑎 − 2 ) ( 𝑎 + 2 ) (a−2)(a+2). Multiplying both sides by 𝑎 − 2 a−2 (assuming 𝑎 ≠ 2 a  =2) leads to the equation − 3 𝑎 + 2 𝑎 + 2 = 1 a+2 −3a+2 ​ =1, which simplifies to 𝑎 = 0 a=0. The value 𝑎 = − 2 a=−2 is not valid because it makes the denominators 𝑎 2 − 4 a 2 −4 and 𝑎 + 2 a+2 equal to zero, resulting in division by zero. Therefore, 𝑎 = − 2 a=−2 is an extraneous solution, and the correct solution to the original equation is 𝑎 = 0 a=0.