single root of a cubic equation

Question

question: Find all the exhaustive values of b for which the equation $$4(1-b)t^3 + 4bt^2 + (b-1)t - b =0$$ possesses a solution.

i thought about applying rolles theorem, but could not move a step forward. Please help.

Answer 1

Express from the given equation $b$ in terms of $t$. For all $t \ne \pm 1/2$ we have $$ b= {t\over t-1} = :f(t)$$

Since the range of $f$ is $\mathbb{R}-\{1\}$ we have at least $b\in\mathbb{R}-\{1\}$

Especially for $b=1$ we get $4t^2-1=$ which also have a solution. So finally we have $b\in\mathbb{R}$

Answer 2

Writing the LHS as $f(t)$, if $b \ne 1$ then $f$ is an honest-to-god cubic polynomial, and you have that the $f(t) \to + \infty$ as $x \to +\infty$ and $f(t) \to -\infty$ as $x \to -\infty$ (if $b<1$, and vice-versa if $b>1$). Consider the first case: then you can find a point $a<0$ and a point $b>0$ such that $f(a)<0$, $f(b)>0$, and $[a,b]$ is not a trivial interval. Therefore by the intermediate value theorem there exists an $x \in [a,b]$ such that $f(x)=0$, which is what you want.

The case $b>1$ is similar - the only thing to check is what happens if $b=1$. In that case we have $f(t) = 4t^{2}-1$, which obviously has a solution. So there are solutions to the equation $f(t)=0$ for all values of $b$.