Question
Let x_1 and x_2 be the roots of the equation x^2 + 3x + 1 = 0. Compute
\Big( \dfrac{x_1}{x_2+1} \Big) ^2+\Big( \dfrac{x_2}{x_1+1} \Big) ^2
Let x_1 and x_2 be the roots of the equation x^2 + 3x + 1 = 0. Compute
\Big( \dfrac{x_1}{x_2+1} \Big) ^2+\Big( \dfrac{x_2}{x_1+1} \Big) ^2
Since x_1 and x_2 are the roots of the equation x^2 + 3x + 1 = 0, we get the following equations by Vieta's formula.
x_1+x_2=-3
and
x_1x_2=1
Then,
x^2_1+x^2_2
=(x_1+x_2)^2-2x_1x_2
=9-2=7
Therefore,
\Big( \dfrac{x_1}{x_2+1} \Big) ^2+\Big( \dfrac{x_2}{x_1+1} \Big) ^2
=\Big( \dfrac{x_1}{x_2+1} +\dfrac{x_2}{x_1+1} \Big) ^2-\dfrac{2x_1x_2}{(x_1+1)(x_2+1)}
=\Big( \dfrac{x^2_1+x^2_2+x_1+x_2}{x_1x_2+x_1+x_2+1}\Big) ^2 -\dfrac{2x_1x_2}{x_1x_2+x_1+x_2+1}
=\Big( \dfrac{7-3}{1-3+1}\Big) ^2-\dfrac{2}{1-3+1}
=18