# How To Solve Quadratic Equations

This chapter I discuss quadratic equations and how they are useful to model nonlinear data.

In previous sections we discussed linear equations and functions. Let us now increase the difficulty ever so slightly to the famed quadratic equations and functions. A quadratic function will have an $$x^2$$ in it.

Something like:

$x^2 + 2x -3$

The domain of a quadratic function is all real numbers. Its graph is a parabola. The parabola can open either up or down. A parabola opens upward if the leading coefficient is positive. Conversely, it opens downward if the leading coefficient is negative.

The highest or lowest point of a parabola curve is the vertex. A line passing through the vertex evenly is the axis of symmetry.

Leading coefficients of the quadratic function also control the width of the parabola. Large values make a parabola narrow while smaller values make the parabola wider.

### Completing the Square

When a quadratic function is in the form $$ax^2 + bx + c$$, you do not know where the vertex is. If you remember, the vertex is the lowest or highest point of the graph.

To know where the vertex is, the equation must be in a different form. It must look like $$a(x-h)^2 +k$$. In this form, the vertex is located at the "h" and "k" positions. This is called the vertex form.

### Vertex Formula

This formula will be very important hereafter. It is $$\frac{-b}{2a}$$. It is key to different types of problems later on. It is also part of the very famous quadratic formula.

### Example

Here is an example I found that should illustrate how this is done.
Convert the vertex form to the quadratic function form.
Change $$f(x) = 2(x-1)^2 +4$$ to the quadratic form.

$2(x-1)^2 + 4 = 2(x^2 - 2x +1) + 4$

$= 2x^2 - 4x +2 +4$

$= 2x^2 - 4x + 6$

We can change the form back the other way around too. Learning to do this is very useful. Depending on what you have to do, manipulating the formula to be in the correct form is essential. We do this by completing the square.

### Example

Convert $$x^2 + 6x - 3$$ to its vertex form.
We do this by completing the square.

$f(x) = x^2 + 6x - 3$

$y + 3 = x^2 + 6x$

So, in this equation, k=6. Add $$\frac{k}{2}^{2}$$ to both sides.

$y + 3 + 9 = x^2 + 6x + 9$

$y + 12 = (x + 3)^2$

$y = (x + 3)^2 - 12$

These types of problems have a nice practical side too. This next example is a classic word problem given by teachers everywhere. We will look at it and go step by step.

### Example

A farmer is fencing a rectangular area using the straight portion of a river as one side of the rectangle. If the farmer had 2400 feet of fencing, find the dimensions of the rectangle that give the maximum area.

So what we know is that the two small sides plus the length equal 2400 feet.

$A = L*W$

$Width + Width + Length = 2400$

$2W + L = 2400$

$L = 2400 - 2W$

$A = L*W$

$A = (2400 - 2w) * W$

$A = 2400W - 2W^2$

$A = -2W^2 + 2400W$

Now we use the vertex formula to find our missing values.
Vertex forumula is $$\frac{-b}{2a}$$.

$W = \frac{-2400}{(2)(-2)}$

$W = 600$

We know from earlier what our length is.

$L = 2400 - 2W$

Now we just replace the value for $$W$$.

$L = 2400 - (2)(600)$

$L = 1200$

So the dimensions for our rectangle are $$600 \text{by} 1200$$.

There I are some things I learned when I first did this. Use your symbols as long as you can before you start entering values in. The reason why is that sometimes you can get off track by worrying about the values for this and that variable.

The other main thing I learned when attempting this was that the vertex formula just wants the coefficients. It does not need the variable or the value for that variable. That wasn't really clear to me at first.

Let us do another problem of a similar type.

### Example

Here is another problem. A farmer has 1000 feet of fence to enclose a rectangular area. What dimensions for the rectangle result in the maximum area enclosed by the fence?

Unlike the last problem, this is a full rectangle. It has 4 sides instead of 3. We approach it the same way.

We know that $$2L + 2W = 1000$$.

So a rectangle has 4 sides. Two of them can be called "L" and 2 can be called "W". That is where the formula comes from.

Our goal is to find both "L" and "W". From the original formula:

$2L = 1000 - 2W$

Divide both sides by 2.

$L = \frac{1000-2W}{2}$

Simplify the right side.

$L = 500 - W$

Remember that $$A = LW$$. That means:

$A = (500 - W) (W)$

Multiply those two expressions together.

$A = -W^2 + 500W$

Now we use our vertex formula, $$\frac{-b}{2a}$$.

$\frac{-500}{(2)(-1)}$

$\frac{-500}{-2}$

$W = 250$

We just need "L" now but we figured out the formula for that earlier:

$L = 500 - W$

Substitute in "W" now.

$L = 500 - 250$

This gives us:

$L = 250$

This this gives us the dimensions "250 x 250".

### The Motion Formula

The motion formula is very popular in both algebra and calculus. It looks like:
$s(t) = -16t^2 + v_0t + h_0$

There are multiple versions of this formula, depending on the units. This
version of the formula calculates the height of the object in feet after "t"
seconds.

The $$h_0$$ portion of the formula denotes the initial height of the object.
$$v_0$$ denotes the initial velocity above some reference point. If the initial
velocity is upward then $$v_0$$ is greater than zero. However, if the intial
velocity is downward then $$v_0$$ is less than zero.

### Example

Here is a problem about the flight of a baseball.
The initial velocity is 80 feet per second.
Initial height after it leaves the bat is 3 feet.
How high is the baseball after 2 seconds?
Lastly, find the maximum height of the baseball.
Remember our motion formula:

$s(t) = -16t^2 + v_0t + h_0$

The problem gives us all the values in a straightforward manner.
We can just substitute them directly into the formula.
Our formula now looks like this:

$s(t) = -16t^2 + 80t + 3$

This is our basic formula with the information we are given.
How high is the baseball after 2 seconds?
So, we already have a time variable in the formula just waiting to be used.
We just plug it in there and do the calculation.

$s(2) = -16(4) + 160 + 3$

$s(2) = 99$

So our baseball attains a high of 99 feet after two seconds.
The next questions is to find the max height of the baseball.
Our "a" coefficient is negative so that means the parabola opens downward.
The vertex will be the highest point on the graph.

$t = -\frac{b}{2a}$

$t = -\frac{80}{-32}$

$t = 2.5$

That gives us the x-coordinate.
We put that back into the formula just like we did before.

$s(2.5) = -16(2.5)^2 + 80(2.5) + 3$

$s(2.5) = 103$

The max height of the baseball reached 103 feet and it took 2.5 seconds to get there.
Thinking about motion in this manner is pretty interesting!
Let's do some examples for practice now to cement our knowledge.

### Example

Is this expression linear or quadratic?
$f(x) = 1-2x + 3x^2$
It needs to be written in the proper form.
$f)x) = 3x^2 -2x +1$
This will be a quadratic function.

### Example

Is this expression linear or quadratic?
$f(x) = \frac{1}{x^2-1}$
Ths expression looks weird right and is nothing like our examples before.
This expression is therefore neither linear or quadratic.

### Example

Is this expression linear or quadratic?
$\frac{1}{2} - \frac{3}{10} x$
Despite being in a fractional form, this does not have an $$x^2$$ term in it.
That will make this a linear expression.

### Example

Is this expression linear or quadratic?
$f(x) = -3x^2 + 9$
This does not have an "x" term but it is no less a quadratic expression.

### Conclusion

In this chapter we have covered quadratic functions and equations. Along with
quadratic functions come the parabolas. Parabolas are the graphing equivalent
of a quadratic function. If the coefficient of a quadratic frunction is
positive then the corresponding parabola will open upward. Conversely, if
the quadratic function is negative then its parabola will open downward.

We also covered the vertex of the parabola including how to find it. The
vertex formula $$-\frac{b}{2a}, f\frac{-b}{2a}$$ is very important and will
remain so later on.