Sunday, 9 October 2016

Quadratic Equations General Solution Prooff

Quadratic Equations

General form

$$ax^2 + bx + c = 0 .....(1)$$

Find general solution


Divide both sides by a \begin{align} x^2 + \frac{b}{a}x + \frac{c}{a} & = 0\\\\ x^2 + \frac{b}{a}x & = - \frac{c}{a} .....(2)\\ \end{align} Now \begin{align} \left(x + \frac{b}{2a}\right)^2 &= x^2 + \frac{b}{a}x + \left(\frac{b^2}{4a^2}\right) \end{align} Rewrite (2) \begin{align} \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} & = - \frac{c}{a}\\\\ \left(x + \frac{b}{2a}\right)^2 &= \frac{b^2}{4a^2} - \frac{c}{a}\\\\ x + \frac{b}{2a} &= \pm\sqrt{\frac{b^2}{4a^2} - \frac{c}{a}}\\\\ x + \frac{b}{2a} & = \pm\sqrt{\frac{b^2-4ac}{4a^2}}\\\\ x + \frac{b}{2a} & = \frac{\pm\sqrt{b^2-4ac}}{2a}\\\\ x & = -\frac{b}{2a} \frac{\pm\sqrt{b^2-4ac}}{2a}\\\\ x & = \frac{-b \pm\sqrt{b^2-4ac}}{2a}\\\\ \end{align}

So the general solution is $$ x = \frac{-b \pm\sqrt{b^2-4ac}}{2a}$$

Sunday, 2 October 2016

Blogger iPhone and LaTeX

I have spent several days working out how to write mathematical equations that can be created and read on Mac, iPad and iPhone. I thought I had a working solution which will be subject of a lengthy blog post later once I have fully tested it. This involved HTML editors, Weebly, Blogger, a couple of iPad apps and a couple of scripts.

I was disappointed this morning to see that although all worked fine on my iPad the LaTeX commands were not translated to readable equations on my iPhone on Blogger posts. Despair soon vanished when I set Blogger for this particular blog to always use desktop mode. Just navigate to the Blogger design page, select template, click on flower button as shown below and select No, use desktop option on pop up window.

 

 

Saturday, 1 October 2016

Square Root Of 2 Is Irrational

Square Root Of 2 Is Irrational

Let $\sqrt{2} = \frac{p}{q}$ where p and q are two positive integers such that $\frac{p}{q}$ have no common factors.
$$ \left( \frac{p}{q}\right)^2=2 \\ \therefore p^2=2q^2 $$ This means that $p^2$ is an even number and so $p$ is an even number and can be written as $2r$. $$ \therefore 4r^2=2q^2 \\ q^2 = 2r^2 \\ $$ By the same argument as above this means $q$ can be written as $2s$ but then $\frac{p}{q}$ would have the common factor of 2, and so contradicting initial statement.

Therefore $\sqrt{2}$ must be an irrational number.