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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
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