Bunuel wrote:
If \(8 < x < 9\), and \(x^2 = (10 – y)(10 + y)\), which of the following is a possible value for y?
(A) –7
(B) –6
(C) 3
(D) 4
(E) 5
If \(8 < x < 9\), then \(8^2 < x^2 < 9^2\).
Simplify to get: \(64 < x^2 < 81\)
Since \(x^2 = (10 – y)(10 + y)\), we can substitute this value into the above inequality to get: \(64 < (10 – y)(10 + y) < 81\)
Expand to get: \(64 < 100 – y^2 < 81\)
Subtract \(100\) from all sides to get: \(-36 < –y^2 < -19\)
Multiply all sides by \(-1\) to get: \(36 > y^2 > 19\)
[since we multiplied all sides of the inequality by a NEGATIVE value, we REVERSED the direction of the inequality symbols]At this point, we can easily see that \(y = 5\) satisfies the inequality \(36 > y^2 > 19\)
Answer: E
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