The maximum possible number of real roots of the equation $x^{\frac{5}{5}}-6 x^2-4 x+5=0$ is
The maximum possible number of real roots of the equation $x^{\frac{5}{5}}-6 x^2-4 x+5=0$ is
$0$
$3$
$4$
$5$
Solution
Let $f(x)=x^2-6 x^2-4 x+5$
$
\Rightarrow f(-x)=-x^5-6 x^2+4 x+5
$
Number of changes of sign in $f(x)$ are 2 and number of changes of sign in $f(-x)$ are 1 .
$\therefore$ By descarte's rule of signs
Maximum number of +ve real roots are 2 and $-\mathrm{ve}$ real roots are 1.
$\therefore$ Maximum possible real roots are 3