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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