Answer
$v\leq54.82$.
Work Step by Step
$d=v+\frac{v^2}{25},$
To find the range of speed Kerry must travel for her stopping distance not to exceed $175ft$.
$v+\frac{v^2}{25}\leq175,$
Multiplying both sides by $25$.
$25v+v^2\leq4375,$ Substracting $4375$ from both sides.
$v^2+25v-4375\leq0$.
To solve for the trinomial using the quadratic formula for the quadratic equation of $ax^2+bx+c, x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.
In this case, $v^2+25v-4375, v=\frac{-25\pm \sqrt{25^2-4\times1\times(-4375)}}{2\times1}=\frac{-25\pm134.63}{2}$.
thus, $v=-79.82$ or $v=54.82,$
$(v-54.82)(v+79.82)\leq0,$
Since we are solving for the speed here only the positive answers are relevant.
Therefore, the range of speed Kerry must travel for her stopping distance not to exceed $175ft$ is $v\leq54.82$.
