Baltic Way 1991 · Problem 8

Algebra

Let $a, b, c, d, e$ be distinct real numbers. Prove that the equation

\begin{aligned} & (x-a)(x-b)(x-c)(x-d) \\ & +(x-a)(x-b)(x-c)(x-e) \\ & +(x-a)(x-b)(x-d)(x-e) \\ & +(x-a)(x-c)(x-d)(x-e) \\ & +(x-b)(x-c)(x-d)(x-e)=0 \end{aligned}

has 4 distinct real solutions.

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Review

Polynomials

Solution

Solution: On the left-hand side of the equation we have the derivative of the function

f(x)=(x-a)(x-b)(x-c)(x-d)(x-e)

which is continuous and has five distinct real roots.