Question 722271

BEWARE: The Algebra.com system is cutting-off some of the expressions I am trying to write.  Not my fault.!

A little bit of help on #1, but you must still study (your textbook) to make the best sense of it:


{{{2m^2 – m – 15 = 0}}}
You can try to factor by using the format, (2m_______)(m________), and look for pairs of numbers which give the product of -15.  Note, according to F.O.I.L. multiplication method, the O and I products must sum to -m.  


What works?  15,1; 3,5; 5,3; 1,15?
The 5,3 combination seems to do something useful.  
(2m_____5)(m_______3).  Now you need to decide which operation in each binomial.


Skipping that and going to quadratic formula and discriminant:
Know that {{{b^2-4ac}}} is the discriminant for {{{ax^2+bx+c}}}.  Applying discriminant for {{{2m^2-m-15}}}, it will be {{{(-1)^2-4*(2)*(-15)=121}}}, POSITIVE 121.  Read your book to know what this means.  How many Real solutions?  



Completing the Square allows you to change this general form equation, {{{2m^2 – m – 15 = 0}}}, into standard form.  Standard form allows you to easily understand the graph of the equation.  The part of the equation, {{{2m^2-m}}}, represents a rectangle.  The given equation does not really need completion of the square because it is found to be factorable to solve for m, but if you wished to have a graph or know more about the graph, then Completing the square may be very useful.  You would first factor the 2 from the rectangular expression {{{2m^2-m}}}, and then ADD and SUBTRACT the square term, which is like a missing square piece obtained from examening the actual rectangle represented by {{{m^2-(1/2)m}}}.  If you would draw the picture, you will find that the term to use is {{{(1/4)^2}}}.  I omit the rest of the process here.  See and study your book.



{{{2m^2-m=2(m^2-(1/2)m)}}}
{{{2(m^2-(1/2)m + 1/16) - (1/16)*2}}},....
... will get you started.