# 15分 Algebraic method

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

• 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)

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)

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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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