F.4 一元二次方程

1 解(9x–14)2+(9x–14)–20 = 0
設y = 9x–14，則
(9x–14)2+(9x–14)–20 = 0
y2+y–20 = 0
(y+5)(y–4) = 0
y = -5 或y = 4
所以，
9x–14 = -5 　或 　　9x–14 = 4
9x = 9　　　　或 　　9x = 18
x= 1　　　　　或 　　x = 2

2 解方程x2+2x=a2=2a??
是x2+2x = a2+2a嗎?
x2+2x = a2+2a
x2–a2 = 2a–2x
( x + a )( x–a ) = 2 ( a–x )
( x + a )( x–a )–2 ( x–a ) = 0
( x–a )( x + a–2 ) = 0
x = a 或 x = –a–2

3 假設二次方程x2–5x+k(1–x)=0的一個根為2。
a)求k的值
用餘式定理:
f(2) = 22–5(2) + k(1–2) = 0
4–10 + k(-1) = 0
–k–6 = 0
–k = 6
k = -6

b)求方程的另一個根
x2–5x+k(1–x) = 0
x2–5x+[-6(1–x)] = 0
x2–5x–6 + 6x = 0
x2 + x –6 = 0
(x+3)(x–2) = 0
x = -3 或 x = 2


4 假設二次方程x2–6x+c = 0的一個根為3+√2。
a)求c的值
用餘式定理:
f(3+√2) = (3+√2)2–6(3+√2)+c = 0
32+2(3)(√2)+(√2)2–18–6√2 + c = 0
9+6√2+2–18–6√2 + c = 0
-7 + c = 0
c = 7

b)求方程的另一個根,並以根式表示答案
x2–6x + c = 0
x2–6x + 7 = 0
(x+1)(x–7) = 0
x = -1 或 x = 7

5)假設二次方程kx2–8x+1=0的兩個根相等。
a)求k的值
∵方程kx2–8x+1 = 0的兩個根相等
∴Δ = 0
(-8)2–4(k)(1) = 0
64–4k = 0
–4k = –64
k = 16

b)由此,解上述方程
kx2–8x+1 = 0
16x2–8x+1 = 0
(4x–1)2 = 0
4x–1 = 0
x = 1/4

1) (9x-14)^2+(9x-14)-20=0
[(9x-14)+5][(9x-14)-4] = 0
9x-14 = -5
9x = 9
x = 1
or
9x-14 = 4
9x = 18
x = 2