## If Ax^{2} + Bx + C = 0, then solve for x

One of the most common sorts of algebraic equations found in physics is the quadratic equation - one where the equation involves an "x squared" as well as (possibly) an x and some other constants.

The simplest quadratics to solve are ones like: x^{2} = A - there are said "to have no linear term" meaning that there is no term that is just "x." In these cases, the answer can be obviously found by applying our rule of "doing whatever we need to do to get the unknown by itself" In this case, simply taking the square root of both sides.

There are, however, a couple of important points that need to be considered. Firstly, the sqaure root of A is not the only correct solution to this equation! Observe:

(-1 * sqrt(A) )^{2} = (-1)^{2} * (sqrt(A))^{2} = 1 * A = A

Thus, the negative of the solution we originally found is also a solution! So, we need to write, for completeness, that x = +/- Sqrt(A). However, the situation in physics (as opposed to mathematics) is made simpler by the fact that we are usually trying to solve for some physical value that has to make sense. If, for example, we are trying to find the time a ball lands after being dropped at time t=0, a negative answer will make absolutely no sense - so we know that we only need the positive answer. If instead, we were looking at a ball that was thrown upwards and we observed that at t=0, it was momentarily at rest (at the top of it's flight) and we wnated to infer the time it was thrown upwards (a question that is solved with exactly the same equation as the previous one) we know that the time we want must be negative, so we take that solution.

Secondly, there is a further complication if A is not a positive number - for as we all know, the square root of a negative number is not a real number - it is an imaginary one! It is important to think about whether an imaginary solution makes any sense at all - in a problem like those above, getting imaginary times makes no sense - either we have made a mistake, or there are simply no times where what ever we were looking for happened - in this case we say that there are "no solutions" (as opposed to saying that there are "imaginary solutions" when such solutions make sense).

The more general quadratic equations – the ones involving an "x" term as well as an "x^^{2}" term are slightly more complicated to solve. There are two main ways of solving these equations – factorizing or using the "quadratic formula."

## Factorizing

All quadratic equations can be factorized into the product of two linear terms. For example:

x^{2} + 3x +2 = (x+1)(x+2)

I’ll leave it as an exercise for you to check that this is correct.

Then, if we have the equation

Then we can equivalently say that

In this case, we can clearly see that if

then the equation becomes:

so x=-1 is a solution.

Similarly, it is clear that x=-2 is also a solution.

In general, if we have an equation of the form,

Then we can rewrite this as

and then factorizing

and x = -D and -E

where

And

Again, I will leave this as an exercise for you to check.

In all cases, it is possible to factorize the equation in this way. However, in many cases, the required D and E are difficult to find from simply looking at the equation (that is how I found the 1 and 2 in the numerical example above.

There are two cases where the factorization is especially easy – they are known as “difference of two squares” and “perfect square.”

## The Difference of Two Squares

In this case, we have a quadratic equation that looks like

In this case, as discussed above, x= +/- Sqrt(c)

We can also find this by factorizing the equation – we need to find two numbers that add to give zero and multiply to give c

It is clear that the only answer is:

(x – sqrt(c))(x + sqrt(c))

which gives the answer we found earlier. Any quadratic that has no "x" term can be factorized this way.

** The perfect square. **

Consider the quadratic equation that is factorized to give

Multiplying this out, we get:

So, anytime we have a function like this, where the constant term’s square root is exactly half the coefficient of the linear term, the quadratic equation is a perfect square and has a "double solution"

In many cases, the fact that this is a double solution (rather than a single solution) is usually unimportant – if you are trying to find a solution, then your situation is even easier – there is only one choice!

** The Quadratic Formula. **

Although every quadratic can be factorized (as long as you are ok with imaginary and complex solutions), it is generally hard to find the factorization by simple inspection of the equation. Fortunately, we have an expression that gives the solutions to any quadratic equation.

then

x= (-b +/- Sqrt(b^2-4ac))/2a

This expression gives the answer to any quadratic equation. It is important to consider the possible types of solutions.

There are three distinct possibilities, each based on the value of the discriminant: b^{2} – 4ac. This value can be positive, negative or zero, giving distinct types of solutions to the quadratic equation (assuming a, b and c are all real numbers).

1) Positive Discriminant: In this case, the square root gives a real value. Then the plus or minus in front of it gives you two distinct solutions to quadratic equations.

2) Zero Discriminant: In this case, we have only one solution. The equation is a perfect square.

3) Negative Discriminant: When we take the square root of a negative, we get imaginary solutions. This leads to a pair of solutions that have the same real part and imaginary parts of opposite sign – complex conjugate solutions.