IF
for real numbers and . If the exhaustive range of values that can take is given by , find
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Solving the 2 equations, we get y = 2 1 ( − − 3 x 2 + 8 x − 4 − a + 4 ) , z = 2 1 ( − 3 x 2 + 8 x − 4 − a + 4 ) or y = 2 1 ( − 3 x 2 + 8 x − 4 − a + 4 ) , z = 2 1 ( − − 3 x 2 + 8 x − 4 − a + 4 )
Hence, in order for y and z to be real, we have the condition that − 3 x 2 + 8 x − 4 − a + 4 ≥ 0
Sketching out the graph for − 3 x 2 + 8 x − 4 − a + 4 we get 3 2 ≤ x ≤ 2
Thus, γ = 3 2 , δ = 2 , so 1 2 ( γ + δ ) = 3 2