Gujarati
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

A

$0$

B

$3$

C

$4$

D

$5$

Solution

(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

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