Tues 19/8/14
Lets end the day with a laugh - Thanks Rosala
http://web2.0calc.com/questions/lets-end-the-day-with-a-laugh_9
Matrices and the web2 calc.
http://web2.0calc.com/questions/how-do-i-calculate-the-inverse-of-a-matrix-using-the-sc#r0
♬ ♬ ♬ MELODY ♬ ♬ ♫♪ You light up my life ♪ ♫
@@ End of Day Wrap : Mon 18/8/14 Sydney, Australia Time 11:15pm ♬
Hello all,
Well I am back! I have travelled 2000km (by car) in the last 5 days and attended a wedding. How about that!
I have been really lucky to have DragonSlayer (the Defender), and Bumblebee (The Queens eyes and ears) caring for camelot while I have been away. Thank you, you have done an excellent job. Thanks also to Chris and Alan, Aziz and Rosala, Ninja, Kitty and Zegroes. Plus anyone else who has helped. I really do not know who has been answering all the questions while I have been preoccupied. I've been on the forum for many hours this afternoon but I did not get very far I am sorry.
Also, now that Mr. Massow has changed the layout it is more difficult for me to collect all the answerer names.
Anyway, I will do the best I can.
If anyone writes or sees an excellent answer or post could you please let me know so that I can ask Ninja to include it in one of his threads.
Ninja, are you about. I don't remember stumbling over any of your posts recently? There are a couple of threads here I would like you to archive.
Here are a couple of threads that I put aside. I am sure there are many others. I have not seen a lot of the posts, again I am sorry.
What to do to combat pop-ups
http://web2.0calc.com/questions/non-math-help
* Statistics - Z scores
* Introductory Algebra
http://web2.0calc.com/questions/5x-21-2x-what-does-x-mean
* Solving quadratic formula
http://web2.0calc.com/questions/help-me-with-solving-this-equation-please
*Linear Representation (co-ordinate geometry) of a worded problem.
Thanks to all of you. I always wanted this forum to be like a club and I think that is the way it is going.
I am proud of it and proud of all of you.
♫♪ ♪ ♫ ♬ ♬ MELODY ♬ ♬ ♫♪ ♪ ♫
6x2 - 5x - 4 factors nicely into (2x + 1)(3x - 4)
$$6x^2-5x-4$$
To factor this you have to look for 2 numbers.
They have to multiply to give 6*-4=-24
Since this is negative, one of the numbers will be negative and the other one is positive.
They have to add to -5 this means the larger digit will be the negative one.
6*4=24 but there is no way these will add or subtract to give -5
2*12=24 but thes can't make -5 either.
8*3=24 and 8-3 is 5 they might be helpful. How about 3-8=-5 excellent and 3*-8=-24 So the numbers are 3 and -8
We are going to split -5 up into -8x+3x like this
$$\\6x^2-5x-4=0\\
6x^2-8x+3x-4=0\\$$
NOW we will factorise in pairs. Like this:
$$\\2x(3x-4)+1(3x-4)=0\qquad so\\
(2x+1)(3x-4)=0\\$$
That is how I would do it.
HOWEVER
If the factors cannot be found easily it is easier to use the quadratic formula.
$$\\6x^2-5x-4=0\\
a=\;6\quad b=\;-5\quad and \quad c=\;-4 \\$$
$$\\x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\
x=\frac{5\pm \sqrt{(-5)^2-4*6*-4}}{2*6}\\\\
x=\frac{5\pm \sqrt{25+96}}{12}\\\\
x=\frac{5\pm \sqrt{121}}{12}\\\\
x=\frac{5\pm 11}{12}\\\\
x=\frac{5+ 11}{12}\qquad or \qquad x=\frac{5- 11}{12}\\\\
x=\frac{16}{12}\qquad or \qquad x=\frac{-6}{12}\\\\
x=1\frac{1}{3}\qquad or \qquad x=\frac{-1}{2}\\\\$$
Quadratic equation links: It would be goof for you to look at these links.
http://web2.0calc.com/questions/grouping-method-and-factorizing-quadratic-equations
Help remembering the quadratic formula.