1) Find the value of k if x2 – 2x + k = 0 has a double real root.
2)If the quadratic equation x2 + 2x + 3k – 2 = 0 has real roots, find the range of possible values of k.
3)Solve the equation x2 + 7x + 1 = 0 using the quadratic formula.
4)Solve the equation 5x2 – x – 2 = 0 using the quadratic formula.
5)Find the discriminant for the equation –3x2 + x – 1 = 0 and determine its nature of roots.
要詳盡解釋，thz
S4 Maths Questions
2009-06-29 4:58 am
回答 (2)
2009-06-29 5:25 am
✔ 最佳答案
1. x2 - 2x + k = 0 has a double real root.So, discriminant = 0
(-2)2 - 4(1)k = 0
4 - 4k = 0
k = 1
2. x2 + 2x + (3k - 2) = 0 has real roots,
So, discriminant >= 0
(2)2 - 4(1)(3k - 2) >= 0
1 - (3k - 2) >= 0
1 - k >= 0
k <= 1
3. x2 + 7x + 1 = 0
x = {-7 +- sqrt[(7)2 - 4(1)(1)]} / 2
= (-7 +- 3sqrt5) / 2
4. 5x2 - x - 2 = 0
x = {-(-1) +- sqrt[(-1)2 - 4(5)(-2)]} / 2(5)
= (1 +- sqrt41) / 10
5. -3x2 + x - 1 = 0
Discriminant = (1)2 - 4(-3)(-1) = -11 < 0
So, there is no real root for the equation.
That is, the two roots of the equation are imaginary.
2009-06-28 21:27:27 補充：
第5條公式是沒有問題的。
參考: Physics king
2009-06-29 5:22 am
1) since there is a double real roo
so delta = 0
( b^2 -4ac= 0)
so (-2)^2 - 4(1)(k) = 0
4- 4k = 0
k=1
2) delta >= 0
2^2 -4(1)(3k- 2) >=0
4-12k+8 >=0
12-12k>=0
k<=1
3)&4) use x= {-b+/- 開方(b^2-4ac)}/2a
5) 條式你sure係 correct?
so delta = 0
( b^2 -4ac= 0)
so (-2)^2 - 4(1)(k) = 0
4- 4k = 0
k=1
2) delta >= 0
2^2 -4(1)(3k- 2) >=0
4-12k+8 >=0
12-12k>=0
k<=1
3)&4) use x= {-b+/- 開方(b^2-4ac)}/2a
5) 條式你sure係 correct?
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