# Solving Equations I-Expressions Involving Power Functions

**Applying the Toolbox—Solving Equations**

**Example 2.5** Solve the equation -3x(x +1) = -(x +1) ( x^{2 } + 2).

**
Guess and plug in method .** The following table gives the outcome of plugging
in

various numbers into the left -hand and right-hand sides of this equation.

Here the value of the left-hand side

equals the right-hand side.

So far, we’ve identified that when x = -1, the left-hand
side of the equation equals the right-hand

side. Are there any other values of x which do the same? How do you know? This
is

a very poor method.

**
Graphing method to estimate solutions. **Instead of guessing which numbers to
plug in, we

could graph y = -3x(x +1) and y = -(x + 1) ( x

^{2}+ 2) on the same set of axes to get a more

complete picture. Finding the values of x which solve the equation is the same as finding

the x- coordinates of the points where the graphs of y = -3x(x +1) and y = -(x + 1) ( x

^{2}+ 2)

intersect (or cross).

Use your graphing calculator to graph y = -3x(x +1) and y = -(x + 1) ( x

^{2}+ 2) on the same

axes. Change your viewing rectangle to make sure you are seeing all the important features

of your graphs. Trace and zoom to estimate the x-coordinates of the points where the two

graphs cross. Are any of your answers close to x = -1? Are there any other solutions?

Questions to ponder over the semester: How do you know that you have found all the

solutions? Is your viewing rectangle large enough? These questions are not so easy to

answer, but you will be able to by the end of the semester.

Using a graphing calculator will only allow you to estimate the solutions. We need to use

algebra to find exact solutions. But using both graphing and algebra in the exercises will

increase your understanding and allow you to check your answers.

**Algebraic method.**We’ll use our toolbox in a series of steps.

Preliminary work. Get all terms which are on the right-hand side of the equation over to

the left by adding or subtracting them to each side. With all your mental powers, resist

the temptation to cancel out anything by dividing. You will only miss possible solutions.

Here we add (x +1)( x

^{2}+ 2) to each side to get

-3x(x +1) + (x + 1)(x

^{2}+ 2) = 0.

Combine all like terms. Deconstruct the left-hand side—play with it! How do you know

when you have what you need? Experience and practice! Our end goal is to turn the left

side into a product of expressions and then use the Zero Law .

One way to deconstruct the left part of the left-hand side is to see it as a product of

x +1and -3x . The right part is the product of x +1 and x

^{2}+ 2. We have identified a

common factor of x +1:

identify the common term, | |

factor it out. |

Now at this stage we pause and check: are there any other
common terms in the leftover

pieces? There are no other common terms but if there were, we would factor them
out in

the same way. Continue the process of identifying common terms and factoring
them

out, until there are no more. Then add or subtract the leftovers. Our equation
now looks

like

(x +1)( x^{2}- 3x + 2) = 0.

This quadratic term can be factored (always check for this), so we end up with

(x +1)(x - 2)(x -1) = 0.

Use the Zero Law. We did this part of the problem in Example 2.4. The Zero Law
says

that the solutions of (x +1)(x - 2)(x -1) = 0 are the same as the solutions of x
+1 = 0,

x - 2 = 0, and x -1 = 0. Thus our final answers are x = -1, 1, 2.

How do these solutions compare to the answers you estimated with your graphing

calculator?

**Example 2.6** Solve the equation

**Graphing Method.** Following the previous example we
see that we want to graph the

left-hand side and the right-hand side on the same set of axes. We can estimate
the

solutions to this equation by tracing and zooming in on the points of
intersection. Try

this in this case. Remember that y = 0 is just the x-axis. What happens near x =
4 ? In

fact, your calculator is confused—it has no idea what to do at x = 4 .

Using our brains, we see that plugging in x = 4 gives a zero denominator. This
means

that the left-hand side of the equation is undefined at x = 4 .

We can graph the left-hand side with our graphing calculators as long as we have
the

calculator avoid x = 4 . To do this, we must graph the function in two
pieces—first with

x values smaller than 4, and then with x values larger than 4.

Step 1: Graph y = 0 and

on the same set of axes using a viewing window with a
maximum x-coordinate of 3.5 or

3.99 (or anything just a bit smaller than 4).

Step 2: Graph these same functions using a window with a minimum x-coordinate of

4.01 or 4.1 (or something just a bit larger than 4).

In each step trace and zoom to estimate the x-coordinates of any place where

crosses the x-axis (which happens to be the second
function y = 0 ).

**Algebraic Method. **Here we already have everything but zero on the
left-hand side.

Since the left-hand side contains algebraic fractions , we want to multiply
through (both

sides of the equation) by the common denominator to get rid of the algebraic
fractions.

In this case the common denominator is (x - 4)^{4} . We now want to
solve this revised

equation.

Any x-value which is a solution of the original equation will be a solution to
this revised

equation. This means that we will find the solutions to the original equation
among those

to the revised equation

Read that last sentence carefully—a very important fact is
that in solving the revised

equation, we may get x-values which are not solutions to the original equation.
So we

have a two-step process: solve the revised equation, then check to see which of
these

values are really solutions to the original equation. This last step is called
checking for

extraneous solutions.

Solve revised equation. Solve the equation

This is one complicated algebraic expression subtracted
from another. When your

equation is of this form, combining like terms will often help simplify it . We
will follow

the form of the examples in the Toolbox.

First notice that there is a factor of both x - 4 and x + 2 in each side of the
minus

symbol. We will take the smallest power of each of these terms as our common
term.

The smallest power of x - 4 is 1 and the smallest power of x + 2 is
. With the

common terms underlined, our equation looks like

We now factor out the common terms and subtract the
leftovers in the following series of

steps.

At this point the left-hand side of the equation is now
the product of three simpler

equations. Use the Zero Law! The solutions to this equation are the solutions to
the three

equations

The first and third are linear equations yielding the
solutions x = 4 and x = 28

respectively. The middle equation can be solved by raising each side to the
power
,

which is the reciprocal of the power
. Thus its solution
is x = -2

Remember our two-step process! These x-values may be the solutions to the
revised

equation, but they may not even be in the domain of the left-hand side function.
We have

to worry about zeros in the denominator as well as negative numbers under a
square root ,

fourth root, etc.

Checking for extraneous solutions. The easiest way to do this is to just start
plugging in

to the original equation the (possibly false) solutions that you have already
obtained.

When you plug in an x-value, do you really get zero on the left? Or must you
conclude

the left-hand side is undefined at that x-value? We see that if we plug in x =
-2 and

x = 28 to the original equation we really do get 0 = 0 . But we have already
seen that

x = 4 gives a zero denominator. So our solutions to the original equation are
only

x = -2,28. How do these check out with the solutions you estimated with the
graphing

method?

Summary : Solving Equations |

**Preliminary work. **Get all terms to the left-hand
side of the equation, leaving only zero

on the right-hand side. Convert all negative exponents into positive exponents.
If you

have algebraic fractions, multiply through by the common denominator, and
continue to

the next step to solve this revised equation.

**Tips on solving equations without algebraic fractions.** It generally helps
to keep

asking yourself the following series of questions in order over and over again.
If you

answer yes to any of them, follow the instructions.

Q1. Basic Questions: Is the left-hand side of the equation
a linear, quadratic, or other

polynomial that can be factored? If yes, …you know what to do. Is the entire
left-hand

side raised to a power? If yes, raise each side to the reciprocal of that power
to

get rid of it, and then repeat question Q1.

Q2. Is the left-hand side of the equation a product of algebraic expressions? If
yes, use

the Zero Law to break the original equation into smaller equations. Repeat the

question process (starting with Q1) to solve each smaller equation.

Q3. Are there any like terms? If yes, combine them following the examples in the

toolbox. Go back to Q2.

If these steps do not lead to a complete solution, you should try anything else
you can

think of. But you might have to rely on the numerical estimates you obtained
from

graphing (or some other source).

**Check for extraneous solutions.** Plug in to the original equation all
solutions obtained

above. Do you get zero in the denominator? Or a negative number under a square
root,

fourth root, etc? Later on in the course, we will see other ways extraneous
solutions will

occur. Any solution which gives any kind of problem like this is NOT really a
solution.

All other solutions are your answers. Check them against the graphing estimates.

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