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If a,b (a≠b), are the real roots of the equation (k + 1)(x^{2 }+ x + 1)^{2 }+ (k  1)(x^{4 }+ x^{2 }+ 1) = 0, k ≠ 1, 0.
Then the product of the roots is
Since the equation (x^{2 }+ x + 1) = 0 . oes not have any real roots, the roots of the original equation will be the root of the equation (kx^{2} + x + k) = 0
Hence product of the roots = k/k = 1
z1 + z2= .
z1 + z2= .
We have to prove that
is true.
Square both sides.
Square both sides again.
2x_{1}x_{2}y_{1}y_{2} ≦ x_{1}^{2}y_{2}^{2 }+ y_{1}^{2}x_{2}^{2} and we get
0 ≦ (y_{1}x_{2}  x_{1}y_{2})^{2}.
It is true because x_{1,} x_{2}, y_{1}, y_{2} are all real.
(a + ib) x (a  ib)
= a^{2}  i^{2}b^{2}
we know i^{2}=1
so, equation becomes a^{2} (1)b^{2} =
a^{2} + b^{2}
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