Real Numbers $r$ and $s$ are the roots of $p(x)=x^3+ax+b$ and $r+4$, $s-3$ are roots of $q(x)=x^3+ax+b+240$. If the sum of all possible values of $b$ is $k$, find $|\frac{k}{5}|$ where |.| stands for greatest integer function.
My Progress :- $\\$
First of all I observed that $p(x)+240=q(x)$; hence $p(r)+240=q(r)$ $\Rightarrow q(r)=240$ similarly $q(s)= 240$. $\\$
$q(r+4)-q(r)=-240$$\\$
$\Rightarrow 3r^2+a=-88$$\\$
$q(s)-q(s-3)=240$$\\$
$\Rightarrow 9s^2+3a=213$$\\$
I can't progress further, I feel that some different approach will be required. Any help will be appreciated. $\\$
Thank You.
It is certain that a cubic having $2$ real roots must, in fact, have $3$. Let the other real root of $p(x)$ be $c$. Then, the roots of $q(x)$ are, since sum of roots is same: $\{r+4, s-3, c-1\}$. We have: $$rcs = -b {\tag 1}$$ $$(r+4)(s-3)(c-1) = -b-240 {\tag 2}$$ $$rc+cs+rs=0 {\tag 3}$$ $$(r+4)(s-3)+(s-3)(c-1)+(r+4)(c-1)=0 {\tag 4}$$ Expanding $(2)$, we get: $$-252=-rs+4sc-3rc+3r-4s-12c{\tag 5}$$ Expanding $(4)$ and using $(3)$ reduces to the equation: $$c=13+4r-3s {\tag 6}$$ Substituting $(6)$ in $(5)$, we get: $$-252=-rs+4s(13+4r-3s)-3r(13+4r-3s)+3r-4s-12(13+4r-3s)$$ Thus, $$12r^2+12s^2-24rs+84r-84s+156=252$$ Which means- $$\left(r-s+\frac 72\right)^2=\frac {81}{4}$$ Thus, $$r-s=1,-8$$ Now we take two cases, firstly, $r=s+1$: Thus, from $(6)$, we get $c=s+17$. Hence, putting in $(1)$: $$s(s+17)(s+1)=-b {\tag 7}$$ Putting in $(2)$- $$(s+5)(s-3)(s+16)=-b-240 {\tag 8}$$ $(7)-(8)$ results in a quadratic equation in $s$, which can be solved easily, and hence corresponding values of $b$ can be found. Similarly take case $2$ where $r=s-8$, substitute in $(1)$ and $(2)$ and subtract to get another quadratic, which can again be solved to get values of $b$.
The computation of the answer has been left to the reader.