標題:
two maths qs
發問:
Two terms, a and b, are inserted btw 2 and 4√2 such that the 4 consecutive terms form a GP. If a,c and b form an arithmetic sequence, then c= The graph of y = x2 + kx + 3 intersects with the line y = x - 1 at P(x1, y1) and Q(x2, y2). y1 + y2 =
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最佳解答:
Two terms, a and b, are inserted btw 2 and 4√2 such that the 4 consecutive terms form a GP. If a,c and b form anarithmetic sequence, then c= 2, a, b, 4√2 form a GP, the common ratio : a/2 = b/a = (4√2)/b a / 2 = b / a b = a2/2 ...... [1] a/2 = (4/√2)/b ab = 8√2 ...... [2] Put [1] into [2] : a(a2/2) = 8√2 a3 = 16√2 a3 = (2√2)3 a = 2√2 Put a = 2√2 into [1] : b = (2√2)2/2 b = 4 a, c, b form an AP, the common difference : c - a = b - c 2c = a + b c = (a + b)/2 ...... [3] Put a =2√2 and b = 4 into [3] : c = (4 + 2√2)/2 c = 2 + √2 ===== The graph of y = x2 + kx + 3 intersects with the line y = x - 1 at P(x1,y1) and Q(x2, y2). y1 + y2 = y = x2 + kx + 3 ...... [1] y = x - 1 ...... [2] From [2] : x = y + 1 ...... [3] Put [3] into [1] : y = (y + 1)2 + k(y + 1) + 3 y = y2 + 2y + 1 + ky + k + 3 y2 + (k + 1)y + (k + 4) = 0 y1 and y2 are the roots of the above equation. Sum of the roots : y1 + y2 = -(k + 1)
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