The least integral value of a for which all the roots of the equation $x^4-4x^3 -8x^2 +a=0$ are real.
let $f(x) =x^4-4x^3 -8x^2 +a=0$
$ f'(x) = 4x^3 -12x^2 -16x $
Put $f'(x) = 0 $ we get x =0, -1, 4. How to proceed further, I am not getting any idea on this. Please guide thanks.
Let $f(x)=x^4-4x^3-8x^2+a$, $$f'(x)=4x^3-12x^2-16x=0 \Rightarrow x=4,-1,0$$ Next $f''(x)=12x(x-2)-16.$ so $f(x)$ has local minima at at $x=4,-1$ and loca max at $x=0.$ Hence $f_{min}=a-128, a-3$, $f_{max}=a$. For four real rootss: $f_{min}<0$ and $f_{max}>0.$ Thus, for four real roots $$a \in( 0,3).$$ Both the integers $a=1,2$ are the solutions.
See the fig. for a=1, below
$f(x)$ for $a=1$">
See the fig. for $a=2$, below
