maths

solve (x-p)(x+p) = x- p

1. if x=p,
then lhs (p-p)(p+p) =0
and rhs p-p=0
both lhs and rhs = 0; the equation is satisfied
Therefore x = p or p=x (root 1)

2. for x <> p
then (x-p)(x+p)/(x-p) = (x-p)/(x-p) = 1
x+p =1
x=1-p or p=1-x (root 2)

slove 3000 = 10t + 50t^2

divide both lhs and rhs by 10, then
300 = t + 5t^2
5t^2 + t -300 = 0

solve for t applying {-b + - sqrt[ sq(b) - 4ac)]}/2a
t = [-1+(1^2-4*5*-300)^(1/2)]/(2*5)
= [-1+sqrt(6001)]/10
~ 7.65 (root1)

or
t = [-1-(1^2-4*5*-300)^(1/2)]/(2*5)
= [-1-sqrt(6001)]/10
~ -7.85(root2)

a.(x-p)(x+p) = x- p
x^2-p^2=x-p
x^2-x-p(p-1)=0
(x+(p-1))(x-p)=0
x=1-p or p

b.3000 = 10t + 50t^2
300=t+5t^2
5t^2+t-300=0
t=(-1±√1^2-4(5)(-300))/2
=(-1±√6001)/2

2006-11-11 18:06:26 補充：
sor. b is wrongt=(-1±√1^2-4(5)(-300))/2(5)=(-1±√6001)/10