Sometimes quadratic expressions are in interesting forms, and may not be obvious as some of the examples I have shown you elsewhere. However, they can be understood and analyzed the same way as other quadratic expressions, using a few techniques to understand them, and then apply the same techniques as used to analyze other quadratic expressions. We shall take a look at some examples.

Unclear Formats

Let’s look at the equation that describes the flight of a ball (Eq. 1).

Eq. 1: \(y = -4.9x(x-2)\)

This is a quadratic expression, as it describes the shape of the parabola that is the flight of the ball (Figure 1). But it doesn’t look like any of the quadratic expression formats we’ve seen earlier. So what do we do with it?

What we can do is turn it into a standard format. Let’s think what we have to do to get that. If we remember what we do with terms placed next to each other in expressions, which is we multiply them together, and we remember what we do when multiplying out brackets, which is to multiply by each term in the brackets separately. This gives us a quadratic expression of (Eq. 2)

Eq. 2: \(y = -4.9x^2+9.8x\)

But this isn’t a quadratic expression is it, because it should have a c term shouldn’t it? Well, it doesn’t have that, it’s just that c = 0 and therefore isn’t shown on the equation, since by convention we don’t show additions, subtractions or multiplications by 0. This gives us the terms a = -4.9, b = -9.8 and c = 0, which can be fed into the quadratic formula to derive the roots of the expression (Eq. 3, Eq. 4)

Eq. 3: \(x={-9.8+\sqrt{9.8^2-4\times-4.9\times0}\over2\times-4.9}=2\)

Eq. 4: \(x={-9.8-\sqrt{9.8^2-4\times-4.9\times0}\over2\times-4.9}=0\)

Another thing to consider is that the original equation was already a quadratic expression in the factored form. This is because the term x could be considered to be (x + 0), which we could then use the FOIL method (Eq. 5) to reach the same expression as (Eq. 2).

Eq. 5: \(-4.9(x+0)(x-2)=-4.9(x^2-2x+0x+0)=-4.9(x^2-2x)\)

Either approach you use, you can derive the vertex form of this equation (Eq. 6), which along with the factored form will give you the information about the flight of the ball.

Eq. 6: \(y = -4.9(x-1)^2+4.9x\)

In this case, using the information provided by the vertex and factored forms, the ball lands 2 seconds after being thrown (the x intercept after the launch time of 0 seconds), and reaches a maximum height of 4.9 m (the y coordinate of the vertex) at 1 second (the x coordinate of the vertex).

Functions of x

Sometimes a quadratic expression might have a function, such as a trig function within it. This may make the expression look confusing, but it is still nevertheless a quadratic expression (Eq. 7).

Eq. 7: \(y = cosx^2+4.5cosx-2.5\)

This is still a quadratic expression, only this time, the cosine of x is in place of the x. If you factor this equation (Eq. 8) you can find where y = 0, the same way as any other quadratic expression, or you can use the quadratic formula (Eq. 9, Eq. 10).

Eq. 8: \(y=(cosx–0.5)(cosx+5)\)

Eq. 9: \(x={-4.5+\sqrt{4.5^2-4\times1\times-2.5}\over2\times1}=0.5\)

Eq. 10: \(x={-4.5-\sqrt{4.5^2-4\times1\times-2.5}\over2\times1}=-5\)

If we wanted to solve for a value of x where y, we’d need to look at the roots of this expression. However we need to consider what the cosine function does to the expression, and what it might tell us about any underlying system described by the expression.

Since the (cosx – 0.5) and (cosx + 5) parts of the equation are the x intercepts of the equation, we can find what the values of x = acos(0.5) and x = acos(-5) are. However, since cosine is a function that returns a value between -1 and 1, we know that the answer can only be x = acos(0.5) (Eq. 11).

Eq. 11: \(x = acos0.5 = 60\)

This gives the solution of x = 60 for this unusual expression.