98, 97, 96

Algebra Level 3

P ( x ) = 2 x 98 + 5 x 97 + 5 x 96 + 5 x 95 + ⋯ + 5 x + 3 = 0 P(x) = 2x^{98} + 5x^{97} + 5x^{96} + 5x^{95} +\cdots + 5x + 3 = 0

If the equation above holds true, which of the following is true?

-2 is a root of P ( x ) = 0 P(x)=0 . − 3 2 -\dfrac 32 is a root of P ( x ) = 0 P(x)=0 . The number of rational roots is 3.

Ojasee Duble by Ojasee Duble , 19, India

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1 solution

Chew-Seong Cheong
May 13, 2018

P ( x ) = 2 x 98 + 5 x 97 + 5 x 96 + 5 x 95 + ⋯ + 5 x + 3 = 5 ∑ k = 0 98 x k − 3 x 98 − 2 = 5 ( x 99 − 1 x − 1 ) − 3 x 98 − 2 = 0 \begin{aligned} P(x) & = 2x^{98}+5x^{97}+5x^{96}+5x^{95}+\cdots + 5x + 3 \\ & = 5\sum_{k=0}^{98} x^k - 3x^{98} - 2 \\ & = 5\left(\frac {x^{99}-1}{x-1}\right) - 3x^{98} - 2 = 0 \end{aligned}

Therefore,

5 ( x 99 − 1 x − 1 ) = 3 x 98 + 2 5 x 99 − 5 = ( 3 x 98 + 2 ) ( x − 1 ) 5 x 99 − 5 = 3 x 99 + 2 x − 3 x 98 − 2 2 x 99 + 3 x 98 − 2 x − 3 = 0 ( x 98 − 1 ) ( 2 x + 3 ) = 0 \begin{aligned} 5\left(\frac {x^{99}-1}{x-1}\right) & = 3x^{98} + 2 \\ 5x^{99}-5 & = (3x^{98} + 2)(x-1) \\ 5x^{99}-5 & = 3x^{99} + 2x - 3x^{98} - 2 \\ 2x^{99} + 3x^{98} - 2x - 3 & = 0 \\ (x^{98}-1)(2x+3) & = 0 \end{aligned}

⟹ x = { − 1 − 3 2 \implies x = \begin{cases} -1 \\ - \frac 32 \end{cases} . Note that x = 1 x=1 is not a solution. There are two rational root and one of them is − 3 2 -\frac 32 .

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@Chew-Seong Cheong , we really liked your comment, and have converted it into a solution. If you subscribe to this solution, you will receive notifications about future comments.

Brilliant Mathematics Staff - 3 years ago

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Thanks a lot.

Chew-Seong Cheong - 3 years ago

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