Breaking roots

Algebra Level 4

How many complex root(s) that satisfy the equation

x 98 99 + x 97 99 + ⋯ + x 2 99 + x 99 + 1 = 0 ? \large \sqrt[99]{x^{98}} + \sqrt[99]{x^{97}} + \cdots + \sqrt[99]{x^2} + \sqrt[99]{x} + 1 = 0?

0 98 3 2 1

Paul Ryan Longhas by Paul Ryan Longhas , 24, Philippines

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1 solution

Paul Ryan Longhas
May 11, 2016

x 98 99 + x 97 99 + . . . + x 2 99 + x 99 + 1 = 0 \sqrt[99]{x^{98}} + \sqrt[99]{x^{97}} + ... + \sqrt[99]{x^2} + \sqrt[99]{x} + 1 = 0

⟹ ( x 99 − 1 ) ( x 98 99 + x 97 99 + . . . + x 2 99 + x 99 + 1 ) = 0 ( x 99 − 1 ) \implies (\sqrt[99]{x} -1)(\sqrt[99]{x^{98}} + \sqrt[99]{x^{97}} + ... + \sqrt[99]{x^2} + \sqrt[99]{x} + 1)= 0 (\sqrt[99]{x} -1)

⟹ x 99 99 − 1 = 0 ⟹ x − 1 = 0 \implies \sqrt[99]{x^{99}} - 1 = 0 \implies x - 1 = 0

Thus, x = 1 x = 1 which not satisfy the equation.

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