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匿名 發問於 科學及數學數學 · 1 十年前

15分 Algebraic method

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.

2 個解答

評分
  • ?
    Lv 7
    1 十年前
    最愛解答

    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

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  • 1 十年前

    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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