practice problems pending


S1.
Let a = sqrt(2014)
=> 4029 = 2a^2 + 1
Equation becomes:
ax^3 - (2a^2 + 1).x^2 + 2 = 0
ax^3 - x^2 - 2a^2.x^2  + 2 = 0
x^2(ax -1) - 2(a^2x^2 - 1) = 0
x^2(ax -1) - 2(ax + 1)(ax - 1) = 0
(ax - 1)(x^2 - 2ax - 2) = 0
x = 1/a is one root.
x^2 - 2ax - 2 = 0 gives
x = [2a +-sqrt(4a^2 + 8)]/2 = a +- sqrt(a^2 + 2)
So the 3 roots are:
a > 1
1/a (middle)
a + sqrt(2 + a^2) (biggest)
a - sqrt(2 + a^2) (smallest since negative)
So x2 = 1/a
Answer = 2

Q2.


S2.
Take 4 to LHS and get:
x/x-3 + x/x-5 ... = x^2 - 11x
=> x = 0 is one solution
=> x/x-3 + x/x-5 ... = x - 11

Now combine 3,19 and 5,17 terms
1/x-3 + 1/x-19 = 2x - 22/(x-3)(x-19)
1/x-5 + 1/x-17 = 2x - 22/(x-5)(x-17)
=> 2(x-11)[1/(x-3)(x-19) + 1/(x-5)(x-17)] = x-11
=> x = 11 is another root.
Now
1/(x-3)(x-19) + 1/(x-5)(x-17) = 1/2
1/(x^2 - 22x + 57) + 1/(x^2 - 22x + 85) = 1/2
Let x^2 - 22x + 57 = u
=> 1/u + 1/(u + 28) = 1/2
=> 2(2u + 28) = u^2 + 28u = 4u + 56
=> u^2 + 24u - 56 = 0
u = [-24 +-sqrt(24*24 + 4*56)]/2  = -12 +- sqrt(144 + 56) = -12 +- 10sqrt(2)

Now when you solve x^2 - 22x + 57 = u
You will get 4 solutions for x.
x = [22 +- sqrt(22^2 - 4(57 - u)]/2
x = 11 +- sqrt(121 - 57 + u) = 11 +- sqrt(64 -12 +-sqrt(200 )
x = 11 +- sqrt(52 +- sqrt(200))
So the biggest solution is 11 + sqrt(52 + sqrt(200))
Answer 263





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