Extra Challenge Solutions #1b)

College Algebra

 

Dr. Robert J. Rapalje

Seminole Community College

Sanford, FL  32773

                                                                            

 

 1b)        Solve the quadratic equation:   .

               Solution: 

                              Of course, as in the previous problem, the easiest way to solve a quadratic equation is to factor it.  But how do you factor a trinomial with numbers like this?   You would probably start with the combination:

                                                               

                                   Notice that the last sign is POSITIVE, so the signs are the same, both negative, and you will be ADDING the OUTER times OUTER and the INNER times INNER to obtain the MIDDLE TERM.  You must find two numbers whose product is, so that if you take  times one of the factors and  times the other factor, and ADD, you get.  Lots of luck doing that!!  However, if you were “lucky” enough to try , and arrange the numbers as follows, this DOES factor!

                                                                

                                                                

                                   The OUTER times OUTER is -55x and the INNER times INNER is -24x, which adds to -79x for the MIDDLE TERM!!

                                   To finish the problem, set each factor equal to zero, and solve:

                                                                

 

                                   Of course, if you have a graphing calculator you can use a program on the TI83+ or TI84 called “Polysmlt”, or with any graphing calculator, you can sketch the graph of   and find the zeros (or x-intercepts) of this graph.  These will be the solutions to the equation.

 

                                   Using algebra techniques, you could always solve using the quadratic formula.

                                      , where

      

       

      

Notice that  just happens to be a perfect square (get out your calculator if you don’t already have it out!!) since .  Since this number within the radical is a perfect square, this confirms indeed that the original problem could have been solved by factoring.  Simplifying the quadratic solution now is not difficult.

   

               or      

                      or      

             or      

                                              

                                   So, which method is easier??  It depends upon how good you are at factoring these difficult triangles, or how good you are with a calculator using the quadratic formula!!

 

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Dr. Robert J. Rapalje Altamonte Springs Campus
Contact me at:   rapaljer@scc-fl.edu
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