If $a, b, c, d$ are the roots of the equation $x^{4} + 2x^{3} + 3x^{2} + 4x + 5 = 0$, then $1 + a^{2} + b^{2} + c^{2} + d^{2}$ is equal to

Using Vieta's formulas for the given quartic equation, the sum of the roots a + b + c + d equals -2 and the sum of the products of the roots taken two at a time equals 3. We use the algebraic identity for the sum of squares, which states (a + b + c + d)^2 = (a^2 + b^2 + c^2 + d^2) + 2(sum of products taken two at a time). Substituting the values gives (-2)^2 = (a^2 + b^2 + c^2 + d^2) + 2(3), which simplifies to 4 = (a^2 + b^2 + c^2 + d^2) + 6. Solving for the sum of the squares yields a^2 + b^2 + c^2 + d^2 = -2, and adding 1 gives the final result of -1.

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