1.10 Equations
in Quadratic Form
College
Algebra: One Step at a Time. Page
148 - 154 #2, 9, 12, 16, 19, 21, 25
NEW
PROBLEMS: #2, 9, 12, 16, 19, 21, 25
Dr. Robert J. Rapalje
Seminole Community College
Sanford, FL 32773
2.
This
is really just a TRINOMIAL,
and as such, it can be factored into the product of two binomials.
Notice that the
FIRST
times FIRST
gives you
which
is 
.
Next, the LAST
times LAST
must give you
,
so try 
,
where the numbers are of opposite sign. In order for the numbers to
subtract and give you a 
for
the middle term, it must be 
Now, you have TWO equations to
solve.
:
or
9.
Notice
that this equation seems to be built in terms of the quantity
. While many people may recommend solving this one by the
Substitution Method, I prefer to do this by what I call the “Straight Out
Factoring Method”. My idea is to recognize that since this is in three
parts as it is, it is really just a TRINOMIAL, and as such, it can be
factored into the product of two binomials. Let’s do some warm-up problems
first:
Factor:
Now:
Notice that this factors as above into the product of two
binomials like this:
Notice
that the
FIRST
times
FIRST
gives you
. Next, the LAST
times
LAST
must give you 
, so try 
.
Next, drop
the parentheses within the brackets, and change the brackets to parentheses
to make the equation look simpler.
Now, you
have TWO equations to solve.
FIRST EQUATION.
The first
equation does NOT factor, so you will have to solve this one by either the
quadratic formula or by completing the square. I recommend the completing
the square method.
Begin by
adding -5 to each side of the equation.
Next, you
need to build a perfect square trinomial on the left side by adding a number
to each side of the equation.
You can
find this number by taking HALF of the
and squaring it. Half of
is 
, and 
is 
, so add 
to each side of the equation.
Take the
square root of each side. Don’t forget the
sign!
Finally,
add 
to each side:
SECOND EQUATION.
The second equation DOES factor, so this one should be easy.
FINAL ANSWER.
There are four answers:
or
This problem can also be solved by the Method of Substitution!
12.
Notice that this equation seems
to be built in terms of the quantity
. It is therefore appropriate to name a new variable (you
can use any letter that you like, so let’s use the letter
), and write 
. This is called the Substitution Method, because you
will be substituting this back into the original problem and write this:
Now, you
can see that this equation is in the shape of a quadratic equation. That is
why we call this section “quadratric form.” Not only is this in the form of
a quadratic equation, it factors.
Now, you
must substitute the formula for , and solve for the original variable which is
. This gives you the following two equations to
solve:
Multiply
both sides of each equation by the common denominator which is
.
Interestingly enough, both of these trinomials factor!
16.
In this equation, the
expression 
is
the reciprocal of 
.
It is therefore appropriate to let
.
This means that the reciprocal
.
When these two substitutions are made into the original equation, it looks
like this:
Next, multiply both sides of
the equation times 
in
order to clear the fraction.
The denominator divides out
leaving:
This factors into:
Now, you must substitute the
formula for ,
and solve for the original variable which is
.
This gives you the following two equations to solve:
Multiply both sides of each
equation by the common denominator which is
.
Final
answer: 
These answers are guaranteed, since you did NOT square both sides. You did
multiply both sides by the variable x, but since
and
,
checking the answers is not required. There will be
NO extraneous roots!!
19.
In this equation, the
expression 
is
the reciprocal of 
.
It is therefore appropriate to let
.
This means that the reciprocal
.
When these two substitutions are made into the original equation, it looks
like this:
Next, multiply both
sides of the equation times in
order to clear the fraction.
The denominator
divides out leaving:
This factors
into: 
Now, you must
substitute the formula for ,
and solve for the original variable which is
.
This gives you the following two equations to solve:
Multiply both sides
of each equation by the common denominator which is
.
The first problem
above can be factored, but the second problem does NOT factor, so it must be
solved by completing the square or by quadratic formula. In this case, I
think completing the square works best. First, the factoring problem:
For the second
problem use the completing the square method:
Take half of the
4
and square which is 4:
Take the square root
of each side:
Final answer:
or
These answers are
guaranteed, since you did NOT square both sides. You did multiply both
sides by the variable x, but since
and,
checking the answers is not required. There will be
NO extraneous roots!!
21.
It
appears that the expression 
is
the building block of this equation. It is therefore appropriate to let
.
Squaring both sides gives you 
.
When these two substitutions are made into the original equation, it looks
like this:
Notice that this
factors:
Now, you must
substitute the formula for ,
and solve for the original variable which is
.
This gives you the following two equations to solve:
Since the
power
means “cube root”, you must cube both sides of the equations to “undo” the
power
and solve for x.
Solve these quadratic equations. Notice that when set equal to zero, they
both factor!! Hmmm!
Final Answer: 
These answers are guaranteed, since you did NOT square both sides. (Cubing
both sides does NOT generate extraneous real solutions!) There will be
NO extraneous roots!!
25.
While this looks at first like a radical equation, in which you might square
both sides in order to eliminate the radical, if you do this, you will have
to square a trinomial, which is a very LARGE equation to solve. Now, if you
happen to have a graphing calculator that can solve large polynomial
equations, maybe this is not such a bad thing after all! Nevertheless,
there is a building block of 
within
this equation. It is therefore appropriate to let
.
Squaring both sides gives you 
.
When these two substitutions are made into the original equation, it looks
like this:
Set this equal to zero, and notice that it factors:
Now, you must substitute the formula for
,
and solve for the original variable which is
.
This gives you the following two equations to solve:
You must now square both sides of the equations to “undo” the square roots
and solve for .
Solve these quadratic equations. Notice that when set equal to zero, one of
them factors, but the other one does not!! Too bad!! Sometimes life comes
out even, sometimes it does not!
(Does not factor! Solve 
by completing the square!) 
For the first problem use the completing the square method:
Take half of the
8
and square which is
16:
Take the square root
of each side:
(approximately
1.65 and -9.65)
Final answer:
or
The check on this is tricky, but since you squared both sides of the
equations, it is necessary! Substituting each value of
back
into the original equation, especially the radical answers obtained by
completing the square, is way too tedious and hard. There must be an easier
way—and there is!
Check:
You can set this equation equal
to zero, giving you 
.
With a graphing calculator, you can sketch the graph of
and
determine the zeros (or roots) of this graph. That is, you can find the
values at which the graph crosses the x-axis. These values will be the
solutions to the equation. This graph clearly has 4 points of intersection,
each of which corresponds to the solution values that were obtained.
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