Try to rearrange $d = 0.039v^2 + 0.1v$ so that one side is $0$ and the other side looks like $av^2 + bv + c$ in terms of $v$.

Once you have $av^2 + bv + c = 0$, use $v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Pay close attention to the signs of $b$ and $c$ when you compute $b^2 - 4ac$.

You will get two expressions for $v$. Since $v$ is a speed, which sign in front of the square root gives a nonnegative value?

In Desmos, type f(v) = 0.039v^2 + 0.1v. This represents the stopping distance $d$ as a function of speed $v$.

Pick a convenient positive value for $d$, such as d = 50, and keep it fixed. You will test each answer choice using this value.

For each option, type a corresponding expression, for example v_A(d) = (-0.1 + sqrt(0.01 + 0.156d))/0.078, v_B(d) = (0.1 + sqrt(0.01 - 0.156d))/0.078, etc. Then evaluate each at your chosen $d$ (e.g., by creating a table or entering v_A(50), v_B(50), and so on).

For each computed value of $v$, plug it into f(v) (e.g., type f(v_A(50)), f(v_B(50)), etc.) and compare the result to your chosen $d$ (such as 50). The expression that produces a $v$ for which f(v) equals your test $d$ and gives a nonnegative speed corresponds to the correct option.

The given model is

To solve for $v$, move $d$ to the other side so the equation is in standard quadratic form $av^2 + bv + c = 0$:

Compare

to $av^2 + bv + c = 0$.

So:

• $a = 0.039$
• $b = 0.1$
• $c = -d$

The quadratic formula for $av^2 + bv + c = 0$ is

Substitute $a = 0.039$, $b = 0.1$, and $c = -d$ into the formula, but keep it symbolic for now:

First compute the denominator:

Now simplify the expression inside the square root (the discriminant):

So the general solutions for $v$ are

The square root $\sqrt{0.01 + 0.156d}$ is at least $0.1$ for all $d \ge 0$, so:

• The root with the minus sign, $-0.1 - \sqrt{0.01 + 0.156d}$, is always negative.
• The root with the plus sign, $-0.1 + \sqrt{0.01 + 0.156d}$, is nonnegative and represents a valid speed.

Therefore, the expression for speed $v$ in terms of stopping distance $d$ is