## A.3: Solving quadratic equations

All the examples so far have been *linear* equations: none of the unknowns were raised to any power^{*}. In physics, there are often situations described by *quadratic* equations, where some unknown quantity is squared.

When the quantity you’re solving for only shows up to the second power, the process is relatively simple: the last step is just to take the square root of both sides of the equation.

#### Example A.2

Solve the equation

\[ h = \frac{1}{2}gt^2 \]

for \(t\).

Start off just like you would in the previous section:

\[

\begin{align*}

h &= \frac{1}{2}gt^2 \\

2h &= gt^2 \\

\frac{2h}{g} &= t^2

\end{align*}

\]

There are actually two possibilities for what \(t\) could be. One is \( t = \sqrt{\frac{2h}{g}} \). But because the square of a negative number is positive, \(t = -\sqrt{\frac{2h}{g}}\) is also a possible solution. We write the solution concisely as

\[ t = \pm\sqrt{\frac{2h}{g}} \]

where the symbol \(\pm\) is read as “plus or minus.”

Most of the time, it is clear from the physics which solution is the accurate description of reality. If \(t\) represents a length of time, for instance, we would use the positive solution because we would be trying to find the span of time that some process takes, which would be a positive value.^{**}

The process is slightly more complicated when the quantity you’re solving for appears in the equation multiple places and is sometimes not squared. In these cases you’ll use the *quadratic formula*:^{***}

If an equation can be written in the form

\[ 0 = Ax^2 + Bx + C \]

then the solution is given by the quadratic formula:

\[ x = \frac{-B \pm \sqrt{B^2 – 4AC}}{2A} \]

The only constraint is \(A\) cannot be zero. (Though if \(A=0\), you probably wouldn’t have though of the equation as quadratic in the first place.)

This “standard” form says zero equals *some number* times \(x^2\), plus *some other number* times \(x\), plus *another number.* Once your equation is written in this form you can identify each of those numbers and apply the quadratic formula.

#### Example A.3

Solve the equation

\[ h = vt + \frac{1}{2}gt^2 \]

for \(t\).

You see the variable you are solving for, \(t\), appears twice. Once time it is squared, and the other time it is not. Start by rearranging the equation so the quadratic formula can be applied:

\[ 0 = \frac{1}{2}gt^2 + vt – h \]

You can now identify \(A = \frac{1}{2}g\), \(B = v\), and \(C = -h\), and apply the quadratic formula:

\[

\begin{align*}

t &= \frac{-B \pm \sqrt{B^2 – 4AC}}{2A} \\

&= \frac{-v \pm \sqrt{v^2 – 4\left(\frac{1}{2}g\right)(-h)}}{2\left(\frac{1}{2}g\right)} \\

&= \frac{-v \pm \sqrt{v^2 + 2gh}}{g}

\end{align*}

\]

^{*}You could say they were raised to the first power, but that doesn’t really do anything: any number raised to the first power is itself \(\left(x^1 = x\right)\)

**There are examples where the negative solution for time is physically reasonable, but those are beyond the scope of this book.

^{***}There are other methods that may be sometimes be used, and may be more efficient. Applying the quadratic formula also works for those cases and I’ll leave the other methods for a math class.