Find the values of m for which the difference between the roots of the equation $\displaystyle 16x^2 = mx - 2501$ is 5.
I don't know how to start.
Let $\displaystyle a$ and $\displaystyle b$ be roots to this equation.
That means that $\displaystyle 16a^2=ma-2501$ and $\displaystyle 16b^2=mb-2501\implies {\color{red}16b^2-mb+2501}=0$.
However, we impose case that $\displaystyle a-b=5\implies a=5+b$.
Therefore,
$\displaystyle \begin{aligned}16(b+5)^2=m(b+5)-2501 & \implies 16(b^2+10b+25)=m(b+5)-2501\\ & \implies 16b^2+160b+400=mb+5m-2501\\ &\implies {\color{red}16b^2-mb+2501}=5m-160b-400\\ &\implies 5m=160b+400\\ &\implies m=32b+80\end{aligned}$
This tells us that the value of $\displaystyle m$ depends on the value of one of the roots.
Does this make sense?
1. The quadratic equation
$\displaystyle ax^2+bx+c=0$ has the solutions
$\displaystyle x_1=-\frac b{2a}+\sqrt{\frac{b^2-4ac}{4a^2}}~\vee~x_2=-\frac b{2a}-\sqrt{\frac{b^2-4ac}{4a^2}}$
2. The difference between the 2 solutions is
$\displaystyle x_1-x_2=2\sqrt{\frac{b^2-4ac}{4a^2}}$
3. Your equation becomes
$\displaystyle 16x^2-mx+2501=0$
where a = 16, b = -m, c = 2501
Therefore you have to solve for m
$\displaystyle 2\sqrt{\frac{m^2-4\cdot 16 \cdot 2501}{4 \cdot 256}} = 5$
I've got $\displaystyle m=-408$