唔識做功課30

a)x² + (k+2)x + k = 0
Sum of roots = a+p = - (k+2)
Product of roots = ap = k
So
1/a + 1/p = - 1/3
(a+p) / (ap) = - 1/3
- (k+2) / k = - 1/3
3(k+2) = k
k = - 3
b)By the result of k = - 3 in part a),
the equation becomes x² + (-3+2)x + (-3) = 0
⇒ x² - x - 3 = 0
⇒x² = x + 3
Then a² = a + 3 and p² = p + 3 since a and p are roots of the equation.
a³ + 4p²
= a(a²) + 4p²
= a(a+3) + 4(p+3)
= a² + 3a + 4p + 12
= a+3 + 3a + 4p + 12
= 4a + 4p + 15
= 4(a+p) + 15
= 4(1) + 15 ...... Sum of roots of x² - x - 3 = 0 is a+p = 1
= 19

2014-10-03 14:22:59 補充：
Let α , β be the reciprocals of the roots of 2x² - 4x + 5 = 0, then
1/α and 1/β are the roots of 2x² - 4x + 5 = 0.

Product of roots = (1/α) (1/β) = 5/2
αβ = 2/5
Sum of roots = 1/α + 1/β = -(-4)/2 = 2
(α+β)/(αβ) = 2
α+β = 2αβ = 4/5

2014-10-03 14:23:08 補充：
The required quadratic equation is x² - (α+β)x + αβ = 0
x² - (4/5)x + 2/5 = 0
5x² - 4x + 2 = 0

Alternatively :
The required quadratic equation is 2(1/x)² - 4(1/x) + 5 = 0
2 - 4x + 5x² = 0

2014-10-03 14:25:50 補充：
新問題請發新題,謝謝!

2014-10-03 14:38:44 補充：
You're welcome!!!

唔好意思,我唔知唔可以在同一題裡再發問,今次唔該晒!!!