XXI Quadratics

Graphs of Quadratic Function

y = x^2

This function's graph is a combination of two parallel curves called a parabola. There is one value of $y$ for each positive and negative pair of value of $x$. The graph has a minimum value of 0 at the origin, which is also the turning point.
Due to its nature the graph can be used to find square roots.

y = -x^2

Same as the graph of the above function. However all values for $y$ are negative. Also the curves are inverted and moving downwards.

y = ax^2

Same as the first function, however $a$ controls the steepness of the curve. The higher its value, the steeper the curve.

y = x^2 +/- a

Same as the first function but the graph is raised or lowered according to the value of $a$

y = (x-a)^2

Displaces the graph $a$ units to the right, or to the left if its value is added instead of subtracted.

y = (x-b)^2 + a

A general function for any possible position on the coordinate plane with any possible steepness. Implies the standard form of

$y = ax^2 + bx + c$

ax^2 + bx + c = 0

Describes the points where the curve cuts the x-axis. They can be read from a plotted graph where $y$ equals $0$.

Quadratic Inequalities

Inequalities such as $x^2-4x+3 > 0$ and $x^2-4x+3 < 0$ define all the values of $x$ above and below all points where $y = 0$. Therefore they can be solved for all points on the parabola that fall into the regions.

On the other hand general inequalities such as $y > x^2-5x+4$ and $y < x^2-5x+4$ describe the regions above and below the curve $y = x^2-5x+4$.

Quadratic Equations

Algebraic techniques can be used to solve such equations with any required degree of accuracy.

For example, an equation such as $(x-1)^2 - 4 = 0$ can be put transformed to find the corresponding values of $x$ where $y = 0$.

$(x-1)^2 - 4 = 0$
$= (x-1)^2 = 4$
$= x-1 = \pm 2$

$x-1 = 2$
$x = 3$

$x-1 = -2$
$x = -1$

So two valid points of the curve are $(3,0)$ and $(-1,0)$.

Solving by Completing the Square

Given the equation $x^2-2x-3 = 0$

1. Remove the constant on the right:
   $x^2-2x = 3$

2. Divide throughout by the coefficient of $x^2$ if it is anything but unity

3. Add a number that produces a complete square on the left side: (Always the square of half the coefficient of $x$)

$x^2-2x+(-1)^2 = 3 + 1$

This factors into: $(x-1)^2=4$

Solving by Factorization

If the product of two factors is $0$, then at least one of the factors has to be $0$ as well.

Therefore factoring an equation can lead to the values of $x$ where $y = 0$.

Given $(x-1)(x-3)= 0$

$x-1 = 0$
$x = 1$

$x-3 = 0$
$x = 3$

This method can be used to solve equations of higher degrees than $x^2$.

General Formula

Given any quadratic $ax^2+bx+c = 0$

$x_{1,2} = \frac{-b\pm\sqrt[2]{b^2-4ac}}{2a}$

Simultaneous equations of the second degree

A term of $xy$ is a term of the second degree much like $x^2$ or $y^2$.

Therefore <?kt xyz ?> is a term of the third degree.

Simultaneous quadratic equations are only solvable in a very narrow range of cases.

One such case is when one of the equations is a linear equation. In this case the approach to solving is to solve the linear equation for one unknown and substitute it into the quadratic equation.

Example

$x - y = 2 \Rarr y = x-2$
$x^2+xy = 60 \Rarr x^2+x(x-2)=60$
$x^2+x^2-2x=60$
$2x^2 - 2x - 60 = 0$

If the resulting equation happens to have a negative value for $a$, it is correct and flipping signs by multiplying by $-1$ is not legal.

Solving Quadratic Inequalities

Factorization allows to solve these inequalities. Given $x^2-6x+8<0$, results in $(x-2)(x-4)<0$ when factorized.

Since the product of the factors is negative, one factor has to come out negative while the other has to be positive.

In case of $(x-2)(x-4) > 0$ they would both have to yield the same sign for the product to be a positive number.

If $x > 2$ then $(x-2) > 0$ On the other hand if $x < 2$ then $(x-2) < 0$.

For $(x-2)(x-4)$ to yield a pair of opposing signs, $x$ has to be a value between $2$ and $4$.

Thus $2 < x < 4$.

The solution for a positive product would be either $x < 2$ or $x > 4$, that is a value that yields the same sign for both factors.

This method also works for inequalities of higher degrees.

Number lines can come handy when solving these inequalities.