#### 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

Collected in the board: Algebraic equation

Steven Zheng posted 3 hours ago

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

Steven Zheng posted 3 hours ago

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