Problem 1

Algebra Level 4

( x 2010 + 1 ) ( 1 + x 2 + x 4 + . . . + x 2008 ) = 2010 x 2009 \big(x^{2010}+1)(1+x^{2}+x^{4}+...+x^{2008})=2010x^{2009}

Let S S denote the set of all real values of x x such that they satisfy the equation above, then the number of elements in S S is?


The answer is 1.

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Keshav Tiwari by Keshav Tiwari , 23, India

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1 solution

Abdelhamid Saadi
Oct 29, 2016

I will use AM-GM to solve this problem: ( x 2010 + 1 ) ( 1 + x 2 + x 4 + . . . + x 2008 ) = 1 + x 2 + x 4 + . . . + x 4018 ≥ 2010 × 1 × x 2 × x 4 × . . . × x 4018 2010 ( A M − G M ) ≥ 2010 × x 0 + 2 + 4 + . . . + 4018 2010 ≥ 2010 × x 2010 × 2009 2010 ≥ 2010 × ∣ x ∣ 2009 ≥ 2010 × x 2009 \begin{matrix} \big(x^{2010}+1)(1+x^{2}+x^{4}+...+x^{2008}) & = & 1+x^{2}+x^{4}+...+x^{4018}\\ & \geq & 2010\times \sqrt [{2010}]{1 \times x^{2}\times x^{4}\times...\times x^{4018}} & \text (AM-GM)\\ & \geq & 2010\times \sqrt [{2010}]{x^{0 + 2 + 4 + ... + 4018}}\\ & \geq & 2010\times\sqrt [{2010}]{x^{2010 \times 2009}}\\ & \geq & 2010\times {|x|}^{2009}\\ & \geq & 2010\times {x}^{2009}\\ \end{matrix}

The equality needs 1 = x 2 = x 4 = . . . = x 4018 1 = x^{2} = x^{4}=...= x^{4018}

Then x is equal to + 1 +1 or to − 1 -1 .

Only + 1 +1 meets the equality

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Nicely Done. : ) :)

Keshav Tiwari - 4 years, 7 months ago

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