A quick intro to quadratics!
First a quadratic is any equation of the form 
where a,b & c are any real numbers, they produces a graph of the shape:
Solving a quadratic, means we are finding the locations where the function equals zero, which is where the function crosses the y-axis, when 
. These locations or solutions are known as the roots of the quadratic. We can solve these equations using several methods, the general method uses the quadratic formula:
(1) 
An example, is the quadratic 
, so we set 
and 
we get:
![\[ \begin{split} x & =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \\ & =\frac{-(-2)\pm \sqrt{(-2)^{2}-4(3)(-5)}}{2(3)} \\ & =\frac{2\pm \sqrt{4-4(-15)}}{6} \\ & =\frac{2\pm \sqrt{4-(-60)}}{6} \\ & =\frac{2\pm \sqrt{4+60}}{6} \\ & =\frac{2\pm \sqrt{64}}{6} \\ & =\frac{2\pm 8}{6} \\ \end{split} \]](https://andy.bettger.net/wp-content/ql-cache/quicklatex.com-52c39f3ccc1f9fc6627db8ef8a258222_l3.svg "Rendered by QuickLaTeX.com")
We have two solutions because of the 
which is a plus and minus sign, solving the plus first:
![\[ \begin{split} x & =\frac{2+8}{6} \\ & =\frac{10}{6} \\ & =5/3 \\ \end{split} \]](https://andy.bettger.net/wp-content/ql-cache/quicklatex.com-b4b8d5fc745c07e6d05a112c1f84eca6_l3.svg "Rendered by QuickLaTeX.com")
And the minus gives:
![\[ \begin{split} x & =\frac{2-8}{6} \\ & =\frac{-6}{6} \\ & =-1 \\ \end{split} \]](https://andy.bettger.net/wp-content/ql-cache/quicklatex.com-550a3486b1158755ee780b29911e69bb_l3.svg "Rendered by QuickLaTeX.com")
So the solutions to 
are 
and 
, we can check this by looking at the graph of the function:
So we can see that the graph of does cross the y-axis at and as we calculated.
The quantity within the square root of the quadratic formula is known as the discriminant, so 
. We can use the discriminant to classify the quadratics roots. When the discriminant is positive the quadratic will have two solutions or roots, when the discriminant is zero there will be a single solution or root, and the graph of the function will touch the y-axis at a single point. When the discriminant is negative, the solution of the quadratic will be complex, due to the fact that the square root is negative, and therefore complex, the graph of the function will not cross the y-axis at all.
Examples are as follows:
In this example the discriminant is 
, which is positive. So the graph crosses the y-axis, there are two solutions and they are both real and distinct numbers, which are and 
This time the discriminant is 
, so there is a single solution or double root at , which is the location that the graph touches the y-axis.
Finally we have a discriminant of 
, so as this is negative, the solution is complex. We can use the complex number 
which has the property 
to help solve these types of equations.
In this example we would do the following:
![\[ \begin{split} x & = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \\ & = \frac{-2 \pm \sqrt{2^{2}-4*1*2}}{2*1} \\ & = \frac{-2 \pm \sqrt{4-8}}{2} \\ & = \frac{-2 \pm \sqrt{-4}}{2} \\ & = \frac{-2 \pm \sqrt{-1*4}}{2} \\ & = \frac{-2 \pm 2\sqrt{-1}}{2} \\ & = \frac{-2 \pm 2i}{2} \\ & = -1 \pm i \end{split} \]](https://andy.bettger.net/wp-content/ql-cache/quicklatex.com-8178baa6bf09a7928ccc2b5cdf2a84bf_l3.svg "Rendered by QuickLaTeX.com")
So with this example we have two solutions 
and 
, these are a pair of complex conjugates, usually written as 
, this does tell us that the function is at its lowest value at
Taking the factored quadratic 
, we know that the solutions are going to be at 
and 
, we can check this by expanding and then using the quadratic formula:
![\[ \begin{split} y & = (2x)(3x)+(2x)(-2)+(3)(3x)+(3)(-2) \\ & = 6x^{2}-4x+9x-6 \\ & = 6x^{2}+5x-6 \end{split} \]](https://andy.bettger.net/wp-content/ql-cache/quicklatex.com-e7cfccd740fc2cbda9d95244ac126266_l3.svg "Rendered by QuickLaTeX.com")
So using the quadratic formula with 
and 
, we get:
![\[ \begin{split} x & =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \\ & =\frac{-5\pm \sqrt{(5)^{2}-4(6)(-6)}}{2(6)} \\ & =\frac{-5\pm \sqrt{25-4(-36)}}{12} \\ & =\frac{-5\pm \sqrt{25-(-144)}}{12} \\ & =\frac{-5\pm \sqrt{25+144}}{12} \\ & =\frac{-5\pm \sqrt{169}}{12} \\ & =\frac{-5\pm 13}{12} \\ \end{split} \]](https://andy.bettger.net/wp-content/ql-cache/quicklatex.com-e04f600d971e91fd948be4ded621192b_l3.svg "Rendered by QuickLaTeX.com")
So we have two solutions, solving the addition first:
![\[ \begin{split} x & =\frac{-5+13}{12} \\ & =\frac{8}{12} \\ & =\frac{2}{3} \\ \end{split} \]](https://andy.bettger.net/wp-content/ql-cache/quicklatex.com-dce5b5a69388aaf7b1198645233145f8_l3.svg "Rendered by QuickLaTeX.com")
And the subtraction gives:
![\[ \begin{split} x & =\frac{-5-13}{12} \\ & =\frac{-18}{12} \\ & =-\frac{3}{2} \\ \end{split} \]](https://andy.bettger.net/wp-content/ql-cache/quicklatex.com-e334c57c45d40b0970e53aba2c969fbb_l3.svg "Rendered by QuickLaTeX.com")
The solutions to 
are and as expected, from the factored form of the quadratic equation.
