Saturday, November 24, 2012

Determine a if the roots of the equation x^2-x-a=0 are x1 and x2 and x1^4+x2^4=1.

The roots of the equation x^2 - x - a = 0
are


x1 = 1 / 2 + sqrt ( 1 +
4a)/2


x2 = 1/2 - sqrt ( 1+ 4a) /
2


Also x1^4 + x2^4 =
1


=> (1 / 2 + sqrt ( 1 + 4a)/2)^4 + (1 / 2 + sqrt (
1 + 4a)/2)^4 = 1


=> 1/16[(1 + sqrt (1+4a))^4 + (1 -
sqrt (1+4a))^4] = 1


now we use a^2 - b^2 =
(a-b)(a+b)


=> ((1 + sqrt (1+4a))^2 + (1 - sqrt
(1+4a))^2)*((1 + sqrt (1+4a))^2 -(1 - sqrt (1+4a))^2) =
16


=> ((1 + sqrt (1+4a))^2 + (1 - sqrt (1+4a))^2)*(1
+ sqrt (1+4a)) + (1 - sqrt (1+4a))(1 + sqrt (1+4a)) - (1 - sqrt (1+4a)) =
16


=> [1 + 1 + 4a + 2*sqrt (1+4a)+1 + 1+4a -
2*sqrt(1+4a)]*[2*2*sqrt (1+4a)] = 16


=> [4 + 8a
]*[4*sqrt (1+4a) = 16


=> (1+ 2a) * sqrt (1+ 4a) =
1


=> (1+ 4a^2 + 4a)(1+ 4a) =
1


=> 1+ 4a + 4a^2 + 16a^3 + 4a + 16a^2 =
1


=> 8a + 20a^2 + 16a^3 =
0


=> 16a^3 + 20a^2 + 8a =
0


=> 4a^3 + 5a^2 + 2a =
0


=> a( 4a^2 + 5a +
2)=0


a1 = 0


a2 = -5/8 +
sqrt(25 - 32)/8


=> a2 = -5/8 +i* sqrt 7 /
8


a3 = -5/8 - sqrt
7/8


Therefore a can be 0 , -5/8 +i* sqrt 7 /
8 and -5/8 - sqrt 7/8

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