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15分 Algebraic method
發問:
有detail solution 者得15分. 1. Jack cycle from A to B at constant speed. if he increase his speed by 3km/h, he will arrive at B 1/2 hour earlier. distance A to B is 30km, find original cycling speed. 2. if simultaneous equations y= x^2+k y=2x have 1 solution only, find x.
1. Jack cycle from A to B at constant speed. if he increase his speed by 3km/h, he will arrive at B 1/2 hour earlier. distance A to B is 30km, find original cycling speed. 設他原來由A至B的速率是 v，回來時的速率為 V + 3，則 去的時間為 (30/v) 回來的時間為30/(v+3) 依題意 (30/v) = 30/(v+3) + 0.5 [全式乘 v(v+3)] 30(v+3) = 30v + 0.5v(v+3) 30v+90 = 30v + 0.5v2+1.5v 0.5v2 + 1.5v - 90 = 0 v2 + 3v – 180 = 0 (v – 12)(v + 15) = 0 所以 v = 12 或 v = -15 (負值不適合) 所以原來的速率為 12km/h 2. if simultaneous equations y= x2+k _____(1) y=2x _____(2) have 1 solution only, find x. 將(2)式代入(1)式 2x = x2 + k x2 – 2x + k =0 若只有一個解，則判別式為零 B2 – 4AC = 0 (-2)2 – 4(1)(k) = 0 4 – 4k = 0 k = 1 代入原式 x2 – 2x + 1 =0 (x – 1)2 = 0 x = 1 [代入(2)式] y = 2x y = 2x1 = 2
其他解答:
1) Let s km/h be the original speed 30/s = [30/(s+3)] + 1/2 30/s = (60+s+3)/[2(s+3)] 60(s+3) = s(s+63) 60s + 180 = s^2 + 63s s^2 + 3s - 180 = 0 (s + 15)(s - 12) = 0 s = -15 (rejected) or s = 12 (km/h) ************************ 2) y = x^2 + k = 2x x^2 - 2x + k = 0 ----------- (&) Delta = (-2)^2 - 4(1)(k) = 0 4 - 4k = 0 k = 1 Put k = 1 into (&) x^2 - 2x + 1 = 0 (x - 1)^2 = 0 x = 1
15分 Algebraic method
發問:
有detail solution 者得15分. 1. Jack cycle from A to B at constant speed. if he increase his speed by 3km/h, he will arrive at B 1/2 hour earlier. distance A to B is 30km, find original cycling speed. 2. if simultaneous equations y= x^2+k y=2x have 1 solution only, find x.
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最佳解答:1. Jack cycle from A to B at constant speed. if he increase his speed by 3km/h, he will arrive at B 1/2 hour earlier. distance A to B is 30km, find original cycling speed. 設他原來由A至B的速率是 v，回來時的速率為 V + 3，則 去的時間為 (30/v) 回來的時間為30/(v+3) 依題意 (30/v) = 30/(v+3) + 0.5 [全式乘 v(v+3)] 30(v+3) = 30v + 0.5v(v+3) 30v+90 = 30v + 0.5v2+1.5v 0.5v2 + 1.5v - 90 = 0 v2 + 3v – 180 = 0 (v – 12)(v + 15) = 0 所以 v = 12 或 v = -15 (負值不適合) 所以原來的速率為 12km/h 2. if simultaneous equations y= x2+k _____(1) y=2x _____(2) have 1 solution only, find x. 將(2)式代入(1)式 2x = x2 + k x2 – 2x + k =0 若只有一個解，則判別式為零 B2 – 4AC = 0 (-2)2 – 4(1)(k) = 0 4 – 4k = 0 k = 1 代入原式 x2 – 2x + 1 =0 (x – 1)2 = 0 x = 1 [代入(2)式] y = 2x y = 2x1 = 2
其他解答:
1) Let s km/h be the original speed 30/s = [30/(s+3)] + 1/2 30/s = (60+s+3)/[2(s+3)] 60(s+3) = s(s+63) 60s + 180 = s^2 + 63s s^2 + 3s - 180 = 0 (s + 15)(s - 12) = 0 s = -15 (rejected) or s = 12 (km/h) ************************ 2) y = x^2 + k = 2x x^2 - 2x + k = 0 ----------- (&) Delta = (-2)^2 - 4(1)(k) = 0 4 - 4k = 0 k = 1 Put k = 1 into (&) x^2 - 2x + 1 = 0 (x - 1)^2 = 0 x = 1
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