ISI2018-MMA-27

Question

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

• A. $1$
• B. $3$
• C. $5$
• D. $7$

Answer 1

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

Comments

This method doesn’t always work though your answer is correct here. Descartes’ Sign rule only gives the maximum number of real positive and real negative, but there doesn't need to be those many roots of the equation. For Eg:

$f(x)=x^7+5x^5+x^3-3x^2+3x-7$

The above function $f(x)$ has 3 sign changes but has only one root i.e. $x=1$. All the other roots are imaginary because the function $f(x)$ is strictly increasing.