(緊急 , 求求你們)問一元二次方程式的問題 - 10分 (只限今日)

1) 2x^2 + 5x + 3 = 0
(2x + 3)(x +1) = 0
2x + 3 = 0 or x + 1 = 0
x = -3/2 or x = -1

2) 3^2 + x + 2 = 0抄錯題??

3) 2x^2 -7 + 7= 0 都係抄錯??

4) 5x^2 + 6x + 3 = 0
b^2 - 4ac < 0
therefore, no real roots

5) x^2 - 10x + 25 = 0
(x - 5)(x - 5) = 0
x = 5 (repeated)

6) (a^2 + b^2)x^2 - 2abx + b^2 = 0
delta < 0
therefore, no real roots
(其實你係咪抄錯題目呢?!?!?!)

7. 問k為何值 , 則x^2 + 2(1+k)x + k^2 = 0 有等根 ?
因為有等根 , 所以b^2 - 4ac = 0
[2(1+k)]^2 - 4(1)(k^2) = 0
4(k^2 + 2k + 1) - 4k^2 = 0
4k^2 + 8k + 4 - 4k^2 = 0
8k = -4
k = -1/2

8.問m為何值 , 則(m+2)x^2 - 2mx + 1 = 0 有等根 ?
因為有等根 , 所以b^2 - 4ac = 0
(-2m)^2 - 4(m+2)(1) = 0
4m^2 - 4m - 8 = 0
m^2 - m - 2 = 0
(m - 2)(m + 1) = 0
m = -2 or m = 1

9. 設(k+3)x^2 - 4x = -3 有兩相等實根 , 試求k之值 ?
因為有兩相等根 , 所以b^2 - 4ac = 0
(-4)^2 - 4(k+3)(3) = 0
16 - 12k - 36 = 0
-12k = 20
k = -5/3

10. 欲使x^2 - 8x + (m+14)為完全平方式 , 則為m何值 ?
因為 [(x)^2 - 2(x)(4) + (4)^2] = x^2 - 8x + 16 = (x - 4)^2
所以 m + 14 = 16
m = 2

11. k為何值時 , 則 2x^2 -3x - 3 + k(x^2 + x + 2) = 0有兩相等實根 ?
2x^2 - 3x - 3 + kx^2 + kx + 2k = 0
x^2(2 + k) + x(k - 3) + (2k - 3) = 0
因為有兩相等實根 , 所以 b^2 - 4ac = 0
(k - 3)^2 - 4(k + 2)(2k - 3) = 0
k^2 - 6k + 9 - 4(2k^2 + k - 6) = 0
k^2 - 6k + 9 - 8k^2 - 4k + 24 = 0
-7k^2 - 10k + 33 = 0
7k^2 + 10k - 33 = 0
k = -10 正負開方(10^2 - 4(7)(-33) / 2(7)
= (-10 - 32) / 14 or (-10 + 32) / 14
= -3 or 11/7

12. 試證(x - a) (x - b) = m^2恆有實根(a , b , m 均為實根)
(x - a)(x - b) = m^2
x^2 - bx - ax + ab = m^2
x^2 - x(b + a) + (ab - m^2) = 0
(a + b)^2 - 4(1)(ab - m^2)
= a^2 + 2ab + b^2 - 4ab + 4m^2
= a^2 - 2ab + b^2 + 4m^2
= (a - b)^2 + 4m^2
因為 (a - b)^2 > 0 and 4m^2 > 0
所以 x - a) (x - b) = m^2恆有實根

13. 設p , q 是實數 , 要令方程式 2x^2 + 2(p + q)x + p^2 + q^2 = 0有實根 ,
則p不能不等於q , 試證之?
p 不能不等於q?? 即係要等於q??? 係咪有問題?!?!