maths 功課問題

1)
設所買橙的數目是x,

所買的蘋果數目便是15 - x

根據題目,
2(15 - x) + 3x = 36
30 - 2x + 3x = 36
x = 6

所買橙的數目是6個

2)

2(x-3)≧3x+7
2x - 6 ≧ 3x+7
-13 ≧ x
x ≦ - 13

能滿足不等式2(x-3)≧3x+7的最大整數值是 -13

3)

有相等實根, △ = 0

(-k)^2 - 4(k)(1) = 0
k^2 - 4k = 0
k(k - 4) =0
k = 0 或 k = 4

由於k > 0, 捨去 k = 0

因此, k = 4

方程為 4 x^2 - 4x + 1 = 0
(2x - 1)^2 = 0
2x - 1 = 0
x = 1/2

4)
a) 判別式 = (-7)^2 - 4(k)(3) = 49 - 12k

b) 有2個相同的實根, △ = 0
49 - 12k = 0
k = 49/12

1.呢題有2個方法
方法1~~~
設橙的數目是x個，蘋果的數目是(15-x)個
2(15 -x) + 3x = 36
30 - 2x + 3x = 36
∴ x = 6

方法2~~~~
設橙的數目是x個，蘋果的數目是y個
2y + 3x = 36 --- (1)
y + x = 15 --- (2)
Sub (2) into (1),
2(15-x) + 3x = 36
∴x = 6
∴橙的數目是6個

2. 2(x-3)≧3x+7
2x -6≧3x+7
-6-7≧3x-2x
-13≧x
x≦-13
∴最大整數值是 -13

3. kx^2 -kx+1=0 and k>0
△=(-k)^2 - 4(k)(1)=0 (因為有相等實根)
k^2-4k=0
k(k-4)=0
∴k=4 or k=0 (rejected,∵ k>0)
∴k的值是4

代入k=2
4x^2-2x+1=0
(2x-1)^2=0
2x-1=0
x = 1/2


4a) kx^2 -7x+3 =0
Δ=(-7)^2 -4k(3) = 49-12k

b)有相同既實根即係話個判別式(discriminant of quadratic equation) = 0
49-12k =0
49=12k
k=49/12