# Thread: Find the values of m?

1. ## Find the values of m?

Find the values of m for which the difference between the roots of the equation $16x^2 = mx - 2501$ is 5.
I don't know how to start.

2. Originally Posted by womatama
Find the values of m for which the difference between the roots of the equation $16x^2 = mx - 2501$ is 5.
I don't know how to start.
Let $a$ and $b$ be roots to this equation.

That means that $16a^2=ma-2501$ and $16b^2=mb-2501\implies {\color{red}16b^2-mb+2501}=0$.

However, we impose case that $a-b=5\implies a=5+b$.

Therefore,

\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 $m$ depends on the value of one of the roots.

Does this make sense?

3. Originally Posted by womatama
Find the values of m for which the difference between the roots of the equation $16x^2 = mx - 2501$ is 5.
I don't know how to start.
$ax^2+bx+c=0$ has the solutions
$x_1=-\frac b{2a}+\sqrt{\frac{b^2-4ac}{4a^2}}~\vee~x_2=-\frac b{2a}-\sqrt{\frac{b^2-4ac}{4a^2}}$

2. The difference between the 2 solutions is

$x_1-x_2=2\sqrt{\frac{b^2-4ac}{4a^2}}$

$16x^2-mx+2501=0$
$2\sqrt{\frac{m^2-4\cdot 16 \cdot 2501}{4 \cdot 256}} = 5$
I've got $m=-408$