數學 A maths 試題

The equation x²-px+p-1=0, where p is a constant, has distinct real roots│α│and 2α. Find the value of p.
│α│+ 2α=p...(1)
│α│(2α)=p-1...(2)
substitute (1) into (2)
│α│(2α)=│α│+ 2α-1
(│α│-1)(2α)=│α│-1
2α=1
α=1/2
p=│α│+ 2α=1/2+1=3/2

x² - px +(p -1) = 0
a=1, b=p, c=p-1

root 相加 = -b/a
│α│+ 2α = p
when α is positive, 3 α = p (即係 α = p /3)
when α is negative, -α =p (即係 α = -p

root 相乘 = c/a
when α is positive, 2 α^2 = p -1
when α is negative, - 2 α^2 = p -1

兩條公式combine……
when α is positive, 2 (p/3)^2 = p -1
(2p^2) / 9 = p -1
2p^2 = 9p - 9
2p^2 - 9p + 9 = 0
p = 3 OR 1.5
when α is negative, -2 ((-p)^2) = p -1
-2p^2 = p -1
-2p^2 - p + 1 =0
p = -1 OR 0.5
therefore, p = 3, 1.5, -1 OR 0.5
(無時間check數，您自己sub番p落去試下，check下對不對！)