Find $a,b$ such that the equation $x^4+3x^3+6x^2+ax+b=0$ has at most 2 real roots.
I tried doing it with calculus even though, it's at the algebra/polynomial chapter... but however I don't get much information using calculus since:
Let $f(x) = x^4+3x^3+6x^2+ax+b$. We get: $f'(x) = 4x^3+9x^2+12x+a$ and $f''(x)=12x^2+18x+12 >0 \forall x\in\mathbb{R}.$ There for $f'(x)$ is strictly increasing from $-\infty$ to $\infty$ so it has $1$ root. Let's call that root $\alpha$ then we get $f'(x)<0\forall x\in(-\infty,\alpha)$ and $f'(x)>0\forall x\in(\alpha,\infty)$, then $f(x)$ is decresing from $\infty$ to $f(\alpha)$ and then increasing from $f(\alpha)$ to $\infty$. so we get $2$ roots only if $f(\alpha)<0$. Then $f$ has at most 2 real roots $\forall a,b\in\mathbb{R}.$
Do you have any idea how to solve this with algebra?
We show that $f$ has always at most two real roots regardless of the value of $a$ and $b$.
Algebra. If the real quartic $f$ has at least $3$ real roots then all its roots are real, say $x_1,x_2,x_3,x_4$. Then $$x_1+x_2+x_3+x_4=-3\quad\mbox{and}\quad x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4=6.$$ Hence $$0\leq x_1^2+x_2^2+x_3^2+x_4^2=(x_1+x_2+x_3+x_4)^2-2(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4)\\=(-3)^2-2\cdot 6=-3.$$ Contradiction.
Calculus. If $f$ has at least $3$ distinct real roots then by the Mean Value theorem $f'$ has at least $2$ distinct real roots and $f''$ has at least $1$ real root. Now for all real $x$, $$f''(x)=12x^2+18x+12=6(2x^2+3x+2)>0$$ because $\Delta=3^2-3\cdot 2\cdot2=-3<0$. Contradiction.