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Number of real solutions of the equation $x^7 + 2x^5 + 3x^3 + 4x = 2018$ is

  1. $1$
  2. $3$
  3. $5$
  4. $7$
in Numerical Ability by Boss (41.3k points) | 32 views

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Applying Descartes' Sign rule on the equation $f(x)$= $x^7 + 2x^5 + 3x^3 + 4x - 2018=0$

The number of sign changes in $f(x)$ = number of positive real roots = 1

$f(-x)$= $-x^7 - 2x^5 - 3x^3 - 4x - 2018=0$

Number of sign changes in $f(-x)$ = number of negative real roots = 0
Hence number of real roots of the equation = 1
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