If 4 @ 2 = 12 and 6 @ 3 = 27, then what is the value of 8 @ 4?

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Accepted Answer

1. Compute: $4	imes2=8$. Add $2^2=4$ to get $12$.

2. Compute: $6	imes3=18$. Add $3^2=9$ to get $27$.

3. **Rule:** $a@b=a\cdot b + b^2 = b(a+b)$.

4. Then $8@4=8\cdot4 + 4^2 = 32+16 = 48$.

FINAL_ANSWER: a

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