A. Maths中, Sum and Product of Roots of Quadratic Equation問題

As α and β are the real roots of the equation x² - (k+1)x + 2k = 0,

Sum of roots = -[-(k+1)]/1
α+β = (k+1) ................... (1)

Product of roots = 2k/1
αβ = 2k ....................... (2)

(α+1)(β+4)=35
αβ + 4α + β + 4 = 35
αβ + 3α + (α + β) = 31
2k + 3α + (k+1) = 31 ....... by (1) and (2)
3k + 3α = 30
3α = 30 - 3k
α = 10 - k
α = -k + 10

So α = -k + 10

Substitute α = -k+10 into (1),
-k+10 + β = k+1
β = 2k-9

Substitute α = -k+10 and β = 2k-9 into (2),
(-k+10)(2k-9) = 2k
-2k² + 9k + 20k - 90 = 2k
-2k² + 27k - 90 = 0
2k² - 27k + 90 = 0
(2k-15)(k-6) = 0
2k-15 = 0 or k-6 = 0
k = 15/2 or k = 6

So k = 15/2 or k = 6.

x^2-(k+1)x+2k=0
Therefore α+β=-[-(k+1)/1]=k+1
αβ=2k/1=2k

(α+1)(β+4)=35
αβ+4α+β+4=35
2k+k+1+3α=31
3k+3α=30
k+α=10
α=-k+10

Sub α=-k+10 into x^2-(k+1)x+2k=0
(-k+10)^2-(k+1)(-k+10)+2k=0
k^2-20k+100+k^2-9k-10+2k=0
2k^2-27k+90=0
(k-15)(k-6)=90
k=15 or k=6