Maximum possible number of real roots of equation {x^5} - 6{x^2}

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4-2.Quadratic Equations and Inequations

hard

The maximum possible number of real roots of equation ${x^5} - 6{x^2} - 4x + 5 = 0$ is

$0$

$3$

$4$

$5$

(b) Let $f(x) = {x^5} – 6{x^2} – 4x + 5 = 0$

Then the number of change of sign in $f(x)$ is $2$ therefore $f(x)$ can have at most two positive real roots.

Now, $f( – x) = – {x^5} – 6{x^4} + 4x + 5 = 0$

Then the number of change of sign is $1$.

Hence $f(x)$can have at most one negative real root. So that total possible number of real roots is $3.$

Standard 11

Mathematics

medium

(IIT-1987)

normal

(KVPY-2014)

hard

(JEE MAIN-2018)

normal

(KVPY-2017)

easy

(AIEEE-2002)