practice problems pending

P1) If x1 and x2 are non-zero roots of

ax^2 + bx + c = 0 and  -ax^2 + bx + c = 0 , then prove that
a.x^2/2 + bx + c = 0  has a root between x1 and x2.

S1)
Plug x1 and x2 in first and second eqns.
Then let P(x) = a.x^2/2 + bx + c = 0
P(x1) = a.x^2/2
P(x2) = -3/2.a.x^2
i.e. they have opposite signs => it has a root between x1 and x2.

P2)

Let P(x) = x^2 + ax + b be a quadratic polynomial in which a and b are integers. Show that there is an integer M such that

P(n).P(n+1) = P(M) ) for any integer ( n ).

S2)
Let P(x) = (x - p)(x - q) where p,q are the roots and p + q = -a and pq = b.
P(n) = (n - p).(n - q)
P(n + 1) = (n + 1 - p).(n + 1 - q)

Essentially we need to show that 
P(n).P(n+1) = (M - p).(M - q) for some integer M.

P(n).P(n + 1) = (n - p).(n - q). (n + 1 - p).(n + 1 - q)
Now combine (n - p).(n + 1 - q).(n - q). (n + 1 - p)

We want to combine like above so that we can use p + q = a and pq = b to get to P(M) form. 
This will give us:
M = n^2 + n(a+1) + b
and RHS = (M - p)(M - q)

M is clearly an integer here.
H.P.

P3)
Show that
a^3/(a-b)(a-c) + b^3/(b-a)(b-c) + c^3/(c-a)(c-b) = a + b + c

S3)
LHS = -a^3(b-c) - b^3(c-a) -c^3(a-b)/(a-b)(b-c)(c-a)
Numerator is cyclic symmetric.
If you think of it as a polynomial in variable 'b', it becomes 0 at b = c => (b-c) is a factor.
Similarly, (a-b) and (c-a) are also factors.
So numerator = something*(a-b)(b-c)(c-a)
That something is a degree 1 expression since in total degree 4 is required.
Assuming symmetry it's k(a+b+c)
So it becomes k(a+b+c)(a-b)(b-c)(c-a).
Compare the coefficient of a^3 and you will see that k = 1
So LHS simply becomes (a+b+c) hence proved.

P4)
If 1/a + 1/b + 1/c = 1/(a+b+c)
then show that
1/a^n + 1/b^n + 1/c^n = 1/(a^n+b^n+c^n) for all odd n.

S4)

1/a + 1/b = 1/(a+b+c) - 1/c = -(a+b)/(a+b+c).c
=> (a+b)[1/ab + 1/(a+b+c).c] = 0
Either a = -b
or
ab+ac+bc+c^2 = 0
=> a(b+c)+c(b+c) = 0

So eventually:
a = -b or b = -c or c = -a

For a = -b
1/a^n + 1/b^n + 1/c^n
becomes
1/c^n = 1/(0+0+c^n)

Similarly for other cases.
H.P.