# Quadratic Equation

The quadratic function has maximum power of equal to :

- .

When equated to zero, the quadratic function becomes the quadratic equation:

- , in which coefficient is non-zero.

The solution of the quadratic equation is:

The above solution to the quadratic is well known to high-school algebra students. The proof of the solution is usually presented to students as completion of the square, not presented here. To me completion of the square seemed too complicated. To help you acquire a new respect for the quadratic, presented below are more ways to solve it.

## Contents

## Solving the quadratic[edit]

**The depressed quadratic**

The depressed cubic leads to a solution of the cubic and the depressed quartic leads to a solution of the quartic. This paragraph shows that the depressed quadratic leads to a solution of the quadratic.

To produce the depressed function, let

where G is the degree of the function. For the quadratic G = 2. Let

Substitute (3) into (1) and expand:

In the depressed quadratic above the coefficient of is and that of is .

Substitute (5) into (3) and the result is the solution in (2).

**p +- q**

Let one value of be and another value of be . Substitute these values into above and expand.

Substitute into

**By observation and elementary deduction**

You have drawn a few curves by hand and suppose that each curve is symmetrical about the vertical line through a minimum point. Given you suppose that .

You have found the stationary point without using calculus. Continue as per **calculus** below.

**By calculus**

The derivative of is which equals at a stationary point.

At the stationary point .

Prove that the function is symmetrical about the vertical line through .

Let

Substitute (12) in (1) and expand:

Let

Substitute (13) in (1) and expand

The expansion is the same for both values of x, (12) and (13). The curve is symmetrical.

If the function equals 0, then

Substitute this value of p in (12) and the result is (2).

**p + qi**

Let where

Substitute this value of into (1) and expand:

Terms containing

From (14), and

Terms without

From (15) and (16):

This method shows the imaginary value coming into existence to help with intermediate calculations and then going away before the end result appears.

## Defining the quadratic[edit]

The quadratic function is usually defined as the familiar polynomial (1). However, the quadratic may be defined in other ways. Some are shown below:

**By three points**

If three points on the curve are known, the familiar polynomial may be deduced. For example, if three points are given and the three points satisfy , the values may be calculated.

The solution of the three equations gives the equation .

If the three points were to satisfy , the equation would be .

**By two points and a slope**

Given two points and and the slope at , calculate .

Slope = 2Ax + B, therefore

The solution of the three equations gives the polynomial .

**By movement of the vertex**

Begin with the basic quadratic .

If has the value , then and the height of above the vertex = .

If we move the vertex to , then the equation becomes .

If has the value , then

and the height of above the vertex = .

The curve and the curve have the same shape.

It's just that the vertex of the former has been moved to the vertex of the latter .

The latter equation expanded becomes .

The example under "two points and a slope" above is .

Therefore

The example may be expressed as

For proof, expand:

**By compliance with the standard equation of the conic section**

The quadratic function can comply with the format: (See [The General Quadratic] below.)

For example, the function can be expressed as:

or:

To express a valid quadratic in this way, both or both must be non-zero.

**By a point and a straight line**

The point is called the **focus** and the line is called the **directrix**. The distance from point to line is non-zero. The quadratic is the locus of a point that is equidistant from both focus and line at all times. When the quadratic is defined in this way, it is usually called a **parabola**.

## The Parabola[edit]

See Figure 1. For simplicity values are defined as shown. Let an arbitrary point on the curve be .

By definition, . This expression expanded gives:

- and slope = .

Let a straight line through the focus intersect the parabola in two points and .

where

- is the slope of line DB in Figure 1.

Characteristics of the Parabola

The parabola is a grab-bag of many interesting facts. We prove **first** that the tangents at and are perpendicular.

The product of and . Therefore, the tangents (lines AB and AD in Figure 1) are perpendicular.

**Second**, we prove that the two tangents intersect on the directrix. Using and :

The coordinate of the point of intersection satisfies both and . Therefore,

is the mid-point between and . Point A in Figure 1 has coordinates

Check our work:

The tangents intersect at . They intersect on the directrix where . See [Tangents perpendicular and oscillating.]

**Third**, we prove that the triangle defined by the three points and is a right triangle.

Slope of line .

Slope of line .

The product of and is . Therefore the two sides are perpendicular and the triangle in Figure 1 is a right triangle.

## More about the Parabola[edit]

In the last section we proved several points about the parabola, beginning with line and moving towards point on the directrix. In this section, we prove the reverse, beginning with point and moving towards line .

Let be any point on the directrix .

Using and any line through is defined as where is the slope of the line.

Let this line intersect the parabola . (In Figure 1, p = 1.)

The above defines the coordinate/s of any line through that intersect/s the parabola. We are interested in the tangent, a line that intersects in only one place. Therefore, the discriminant is 0.

where

Slope of tangent1 = (In Figure 1, tangent1 is the line .)

Slope of tangent2 = (In Figure 1, tangent2 is the line .)

Prove that tangent1 and tangent2 are perpendicular.

The product of the two slopes is -1. Therefore, the two tangents are perpendicular.

From (24), we chose a value of that made the discriminant 0. Therefore

(In Figure 1, is the slope of line . This statement agrees with proved in the last section.)

We have a line joining the two points . Calculate the intercept on the axis.

Using ,

The line joining the two points passes through the focus .

Two lines parallel

In Figure 2 tangents and intersect at point on the directrix.

The line has value . The line is tangent to the curve at .

The slope of tangent . The slope of line also .

Therefore two lines are parallel.

Area DGBHFD

where

Line . The integral of this value .

Area under line

Area under curve

Area

Similarly it can be shown that Area

Therefore line splits area into two halves equal by area.

## Reflectivity of the Parabola[edit]

See Figure 1. is a right triangle and point is the midpoint of line .

is congruent with , and .

.

1. Any ray of light emanating from a point source at F touches the parabola at B and is reflected away from B on a line that is always perpendicular to the directrix. is the angle of incidence and is the angle of reflection.

2. The path from through focus to vertex and back to focus has length . The path from to to has length all paths to and from focus have the same length.

## Using the Quadratic[edit]

**In Theory**

**1)** The quadratic may be used to examine itself.

Let a quadratic equation be: .

Let the equation of a line be: .

Let the line intersect the quadratic at .

Therefore:

Let the line intersect the curve in exactly one place. Therefore must have exactly one value and the discriminant is .

A line that touches the curve at has slope .

Therefore the slope of the curve at is . This examination of the curve has produced the slope of the curve without using calculus.

Consider the curve: . The aim is to calculate the slope of the curve at an arbitrary point .

If is to have exactly one value, discriminant .

Therefore

The slope of the curve at an arbitrary point .

For more information see [earlier version of "Using the Quadratic."]

**2)** The quadratic may be used to examine other curves, for example, the circle.

Define a circle of radius 5 at the origin:

Move the circle to

We want to know the values of that contain the circle, that is, the values of for each of which there is only one value of .

Put the equation of the circle into a quadratic in .

There is exactly one value of if the discriminant is . Therefore

These values of make sense because we expect the values of to be . This process has calculated a minimum point and a maximum point without calculus.

**3)** The formula remains valid for and/or equal to . Under these conditions you probably won't need the formula. For example can be factored by inspection as .

**4)** The quadratic can be used to solve functions of higher order.

One of the solutions of the cubic depends on the solution of a sextic in the form . This is the quadratic where .

The cubic function produces the depressed function .

The quadratic is solved as .

The roots of the depressed function are .

Using

In this example the quadratic, as part of the depressed function, simplified the solution of the cubic.

The quartic function produces the depressed function which is the quadratic where .

**5)** The quadratic appears in Newton's Laws of Motion:

## The General Quadratic[edit]

Let the directrix be where at least one of is non-zero.

Let the focus be .

Let be any point on the curve.

Distance from point to focus = .

Distance from point to directrix ()

= where .

By definition these two lengths are equal:

.

.

Square both sides:

.

.

.

Expand and the result is:

.

has the form where

If then

.

.

.

Let

Directrix has equation: or .

Focus has coordinates .

becomes . See Figure 1.

Let

Directrix has equation: or .

Focus has coordinates .

becomes or . See Figure 3.

Let

Directrix has equation: or .

Focus has coordinates .

becomes or . See Figure 4.

An interesting situation occurs if the focus is on the directrix. Consider the directrix:

and the focus which is on the directrix.

becomes:

This seems to be the equation of a parabola because , but look closely.

The result is a line through the focus and normal to the directrix.

## Reverse-Engineering the Parabola[edit]

Given a parabola in form the aim is to produce the directrix and the focus.

We will solve the example shown in Figure 5: ,

where:

.

Method 1. By analytical geometry

Find two tangents that intersect at a right angle. The simplest to find are those that are parallel to the axes.

Put the equation of the parabola in the form of a quadratic in .

At the tangent there is exactly one value of . Therefore the discriminant must be .

In the general parabola therefore .

In this example .

Point has coordinates .

The line is tangent to the curve at and has equation: .

Put the equation of the parabola in the form of a quadratic in :

.

By using calculations similar to the above, and .

Point has coordinates .

The line is tangent to the curve at and has equation: .

Point at the intersection of the two tangents has coordinates , and point is on the directrix, the equation of which is: .

Put known values into the equation of the directrix: .

Therefore and the equation of the directrix is: .

The coordinates of points are known. Therefore line is defined as: .

Draw the line perpendicular to . The line is defined as: .

Point at the intersection of lines is the focus with coordinates .

Method 2. By algebra

After rearranging the above values, there are three equations to be solved for three unknowns: :

The solutions are:

If is , the following may be used:

## Area enclosed between parabola and chord[edit]

See Figure 6. The curve is: . Integral is: .

Area under curve

Area under curve area area

Area between chord and curve .

Consider the chord . Call this the with value . The tangent through the origin is parallel to , and the perpendicular distance between is . Call this distance the with value .

In this case the area enclosed between chord and curve the same as that calculated earlier.

Consider the chord and curve By inspection the area between chord and curve

Chord has equation in normal form.

The line is parallel to and touches the curve at

Distance from to chord

Length of

Area between chord and curve the same as that calculated earlier.

These observations suggest that the area enclosed between chord and curve

We prove this identity for the general case. See Figure 7.

Slope of chord

Find equation of chord

Equation of chord

Find equation of tangent

We choose a value of that gives exactly one value.

Therefore discriminant

Equation of tangent in normal form:

Equation of chord in normal form:

Therefore distance between chord and tangent

where

Length of chord

Area under chord

Area under curve

Area between chord and curve

The aim is to prove that:

or

or

or

where:

Because contain a square root, we prove that the squares are equal:

where:

If you choose to make the substitutions and go through all the calculations, you will find that the squares are equal.

Therefore || = || and area enclosed between curve and chord =

where is the length of the chord, and is the perpendicular distance between chord and tangent parallel to chord.