Alg Solving Solutions

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COLLEGE ALGEBRASolutions to Practice Problems

Solving Equations and Inequalities

Paul Dawkins


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College Algebra

Solving Equations and Inequalities

Solutions and Solution Sets

1. Is 6x= a solution to ( )2 5 3 1 22x x = + ?

SolutionThere really isnt all that much to do for these kinds of problems. All we need to do is plug the givennumber into both sides of the equation and check to see if the right and left side are the same value.

Here is that work for this particular problem.

( ) ( )?

2 6 5 3 1 6 22

7 7 OK

= +

=

So, we can see that the right and left sides are the same and so we know that 6x= is a solution to theequation.

2. Is 7t= a solution to 2 3 10 4 8t t t+ = + ?



( ) ( ) ( )?

27 3 7 10 4 8 7

60 60 OK

+ = +

=

So, we can see that the right and left sides are the same and so we know that 7t= is a solution to the

equation.

3. Is 3t= a solution to 2 3 10 4 8t t t+ = + ?


2007 Paul Dawkins 2 http://tutorial.math.lamar.edu/terms.aspx


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College Algebra


( ) ( ) ( )?

23 3 3 10 4 8 3

10 20 NOT OK

+ = +

=

So, we can see that the right and left sides are the not the same and so we know that 3t= is not asolution to the equation.

4. Is 2w= a solution to2 8 12

02

w w

w

+ +=

+?


Note that for this problem we dont even really need to plug the value into the equation. We can see by a

quick inspection that if we were to plug 2w= into this equation we would have division by zero andwe know that is not allowed.

Therefore, 2w= is not a solution to this equation.

Be very careful with this kind of problem. If we had plugged 2w= into the equation wed have gottenzero in the numerator as well and we might be tempted to say that it is a solution to the equation. Wedbe wrong however. Regardless of the value of the numerator, we would still have division by zero and

that is just not allowed and so 2w= will not be a solution to this equation.

5. Is 4z= a solution to 2 26 3z z z + ?

SolutionThere really isnt all that much to do for these kinds of problems. All we need to do is plug the givennumber into both sides of the inequality and check to see if the inequality is true. In this case that willmean checking to see if the left side is larger than or equal to the right side.


( ) ( ) ( )?

2 26 4 4 4 3

8

+

19 NOT OK

So, we can see that the left side is neither larger than nor equal to the right side and so 4z= is not asolution to this inequality.



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6. Is 0y= a solution to ( ) ( ) ( )2 7 1 4 1 3 4 10y y y+ < + + + ?

SolutionThere really isnt all that much to do for these kinds of problems. All we need to do is plug the givennumber into both sides of the inequality and check to see if the inequality is true. In this case that willmean checking to see if the left side is less than the right side.


( ) ( ) ( )( )?

2 0 7 1 4 0 1 3 4 0 10

13 34 OK

+ < + + +

+ ?

SolutionThere really isnt all that much to do for these kinds of problems. All we need to do is plug the givennumber into both sides of the inequality and check to see if the inequality is true. In this case that willmean checking to see if the left side is greater than the right side.


( ) ( )?

21 1 3 1 1

4

+ > +

> 4 NOT OK

Be very careful with this type of problem! Four is not greater than 4 (its equal to 4 big difference here)

and so, we can see that 1x= is not a solution to this inequality.

Contrast the inequality in this problem with,

( )

21 3 1x x+ +

While 1x= is not a solution to the inequality in the problem statement it is a solution to this inequalitysince 4 is in fact greater than or equal to4. The presence of the equal sign in the inequality can make allthe difference in the world and we really need to be on the lookout for it. It is easy to miss when its thereand it is easy to sometimes assume it is there when in fact it isnt.



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Linear Equations

1. Solve the following equation and check your answer.

( )4 7 2 3 2x x x = +

Step 1First we need to clear out the parenthesis on the left side and then simplify the left side.

( )4 7 2 3 2

4 14 7 3 2

11 14 3 2

x x x

x x x

x x

= +

+ = +

= +

Step 2

Now we can subtract 3xand add 14 to both sides to get all the xs on one side and the termswithout an xon the other side.

11 14 3 2

8 16

x x

x

= +

=

Step 3Finally, all we need to do is divide both sides by the coefficient of the x(i.e.the 8) to get the solution of

2x= .

Step 4Now all we need to do is check our answer from Step 3 and verify that it is a solution to the equation. Itis important when doing this step to verify by plugging the solution from Step 3 into the equation given inthe problem statement.

Here is the verification work.

( ) ( ) ( )?

4 2 7 2 2 3 2 2

8 8 OK

= +

=

So, we can see that our solution from Step 3 is in fact the solution to the equation.


( ) ( )2 3 10 6 32 3w w+ =



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Step 1First we need to clear out the parenthesis on each side and then simplify each side.

( ) ( )2 3 10 6 32 3

2 6 10 192 182 4 192 18

w w

w ww w

+ =

+ = =

Step 2

Now we can add 18wand 4 to both sides to get all the ws on one side and the terms without anwon the other side.

2 4 192 18

20 196

w w

w

=

=

Step 3

Finally, all we need to do is divide both sides by the coefficient of the w(i.e.the 20) to get the solution of196 49

20 5w= = .

Dont get excited about solutions that are fractions. They happen more often than people tend to realize.



?

?

49 492 3 10 6 32 3

5 5

64 132 10 6

5 5

78 78OK

5 5

+ =

=

=



4 2 3 5

3 4 6

z z=



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College Algebra

Step 1The first step here is to multiply both sides by the LCD, which happens to be 12 for this problem.

( ) ( ) ( )

4 2 3 512 12

3 4 6

4 2 3 5

12 12 123 4 6

4 4 2 3 3 2 5

z z

z z

z z

=

=

=

Step 2

Now we need to find the solution and so all we need to do is go through the same process that weused in the first two practice problems. Here is that work.

( ) ( ) ( )4 4 2 3 3 2 5

16 8 9 10

2 7

7

2

z z

z z

z

z

=

=

=

=



( ) ( )

7 7?

2 2

35?2

?

4 2 53

3 4 6

4 7 3

3 4 6

11 3 35

3 4 12

11 11OK

3 3

=

+=

= +

=


Note that the verification work can often be quite messy so dont get excited about it when it does.Verification is an important step to always remember for these kinds of problems. You should alwaysknow if you got the answer correct before you check the answers and/or your instructor grades theproblem!




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2

4 1

25 5

t

t t=

Hint : Do not forget to watch out for values of tthat well need to avoid!

Step 1Lets first factor the denominator on the left side so we can identify the LCD. While we are at it we willalso factor a minus out of the denominator on the right side.

( ) ( ) ( )

( ) ( )

2

4 1

25 5

4 1

5 5 5

4 1

5 5 5

t

t t

t

t t t

t

t t t

=

= +

= +

So, after factoring the left side and factoring the minus sign out of the denominator on the right side wecan quickly see that the LCD for this equation is,

( ) ( )5 5t t +

From this we can also see that well need to avoid 5t= and 5t= . Remember that we have to avoiddivision by zero and we will clearly get division by zero with each of these values of t.

Step 2

Next we need to do find the solution. To get the solution well need to multiply both sides by theLCD and the go through the same process we used in the first couple of practice problems. Hereis that work.

( )( )( ) ( )

( )( )

( )

4 15 5 5 5

5 5 5

4 5

4 5

5 5

1

tt t t t

t t t

t t

t t

t

t

+ = + +

= +

=

=

=

Step 3Finally we need to verify that our answer from Step 2 is in fact a solution.

The first thing to note is that it is not one of the values of tthat we need to avoid. Having determined thatwe know that we do have a potential solution (i.e.its not a value of twe need to avoid) all we need to dois plug the solution into the equation given in the problem statement.



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( )

( ) ( )

?

2

?

4 1 1

5 11 25

4 1

1 25 5 1

1 1OK

6 6

=

= +

=



3 4 721 1

yy y

+= +

Hint : Do not forget to watch out for values of ythat well need to avoid!

Step 1

First we can see that the LCD for this equation is,

1y

From this we can also see that well need to avoid 1y= . Remember that we have to avoid division by

zero and we will clearly get division by zero with this value ofy.

Step 2


( ) ( )

( )

3 4 71 2 1

1 1

3 4 2 1 7

3 4 2 51

yy y

y y

y y

y yy

+ = +

+ = +

+ = +

=

Step 3Finally we need to verify that our answer from Step 2 is in fact a solution and in this case there isnt a lotof work to that process. We can see that our potential solution from Step 2 is in fact the value ofythat weneed to avoid and so this equation has no solution.



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We could also see this if we plugged the value of yfrom Step 2 into the equation given in the problemstatement. Had we done that we would have gotten a division by zero in two of the terms! That, of

course, is why we needed to avoid 1y= .

Note as well that we only have caught the division by zero if we verify by plugging into the equation inthe problem statement. Had we checked in the equation we got by multiplying by the LCD it would haveappeared to be a solution! This is the reason that we need to always check in the equation from theproblem statement.


5 6 5

3 3 2 3

x

x x+ =

+

Hint : Do not forget to watch out for values of xthat well need to avoid!

Step 1Lets first factor a 3 out of the denominator of the first tem on the left side so we can identify the LCD.

( )5 6 5

3 1 2 3

x

x x+ =

+

So, after factoring doing the factoring on the first term we can quickly see that the LCD for this equationis,

( ) ( )3 1 2x x +

From this we can also see that well need to avoid 1x= and 2x= . Remember that we have to avoiddivision by zero and we will clearly get division by zero with each of these values of x.

Step 2




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( ) ( )( )

( ) ( )

( ) ( ) ( ) ( )( )

( ) ( )2 2

2 2

5 6 53 1 2 3 1 2

3 1 2 3

5 2 3 1 6 5 1 2

5 10 18 1 5 2

5 10 18 18 5 5 10

28 18 5 10

23 8

8

23

xx x x x

x x

x x x x x

x x x x x

x x x x x

x x

x

x

+ + = + +

+ + = +

+ + = +

+ + = +

=

=

=

Step 3Finally we need to verify that our answer from Step 2 is in fact a solution.

The first thing to note is that it is not one of the values of xthat we need to avoid. Having determined thatwe know that we do have a potential solution (i.e.its not a value of xwe need to avoid) all we need to dois plug the solution into the equation given in the problem statement.


( )( ) ( )

8 ?23

8 823 23

40 ?23

45 5423 23

?

5 6 5

3 3 2 3

6 5

3

8 23 59 9 3

5 5OK

3 3

+ = +

+ =

+ =

=


With this problem we have seen that both the solution and the verification step can be somewhat messy.That will happen on occasion and we shouldnt get excited about it when it does. It is just the way theseproblems work on occasion.

Application of Linear Equations



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College Algebra

1. A widget is being sold in a store for $135.40 and has been marked up 7%. How much did the store payfor the widget?

Step 1Well start by letting pbe the price that the store paid for the widget. The widget has been marked up by7% and so 0.07phas been added to the original price, p.

The equation we get for this problem is then,

0.07 135.4

1.07 135.4

p p

p

+ =

=

Step 2

To finish all we need to do is divide both sides by 1.07 to get the price the store paid for thewidget.

126.5421p=

So, with rounding the store paid $126.54 for the widget.

2. A store is having a 30% off sale and one item is now being sold for $9.95. What was the original priceof the item?

Step 1Well start by letting pbe the original price of the item. The price of the item has been reduced by 30%and so 0.30phas been subtracted from the original price,p.

The equation we get for this problem is then,

0.3 9.95

0.7 9.95

p p

p

=

=

Step 2

To finish all we need to do is divide both sides by 0.7 to get the original price of the item.

14.2143p=

So, with rounding the original price of the item was $14.21.

3. Two planes start out 2800 km apart and move towards each other meeting after 3.5 hours. One planeflies at 75 km/hour slower than the other plane. What was the speed of each plane?

Step 1Lets start with a diagram of what is going on in this situation.



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Step 2

We can next set up a word equation for this situation.

Distance Distance2800

of Plane A of Plane B

+ =

We know that Distance = Rate X Time so this gives to following word equation.

Rate of Time of Rate of Time of 2800Plane A Plane A Plane B Plane B

+ =

Step 3

Lets let rbe the speed of the faster plane. Therefore, the speed of the slower plane is 75r . We alsoknow that each plane travels for 3.5 hours. Plugging all this information into the word equation abovegives the following equation.

( )( ) ( ) ( )

( )

3.5 75 3.5 2800

3.5 3.5 75 2800

r r

r r

+ =

+ =

Step 4

Now we can solve this equation for the speed of the faster plane.

( )3.5 3.5 75 2800

7 262.5 2800

7 3062.5

437.5

r r

r

r

r

+ =

=

=

=

So, the faster plane is traveling at 437.5 km/hour while the slower plane is traveling at 362.5 km/hour (75km/hour slower than faster plane).

4. Mike starts out 35 feet in front of Kim and they both start moving towards the right at the same time.Mike moves at 2 ft/sec while Kim moves at 3.4 ft/sec. How long will it take for Kim to catch up withMike?

Step 1



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Lets start with a diagram of what is going on in this situation.

Step 2

We can next set up a word equation for this situation.

Distance Distance35

Kim Moved Mike Moved

= +

We know that Distance = Rate X Time so this gives to following word equation.

Rate of Time of Rate of Time of 35

Kim Kim Mike Mike

= +

Step 3Both Kim and Mike move for the same amount of time so lets call that t. We also know that Kim movesat 3.4 ft/sec while Mike moves at 2 ft/sec. Plugging all this information into the word equation abovegives the following equation.

( ) ( ) ( ) ( )3.4 35 2

3.4 35 2

t t

t t

= +

= +

Step 4

Now we can solve this equation for the time they traveled.

3.4 35 2

1.4 35

25

t t

t

t

= +

=

=

So, Kim will catch up with Mike after she moves for 25 seconds.

5. A pump can empty a pool in 7 hours and a different pump can empty the same pool in 12 hours. Howlong does it take for both pumps working together to empty the pool?

Step 1



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College Algebra

So, if we consider empting the pool to be one job we have the following word equationdescribing both pumps working to empty the pool.

Portion of job done Portion of job done1 Job

by first pump by second pump

+ =

We know that Portion of Job = Work Rate X Work Time so this gives the following word equation.

Work Rate of Work Time Work Rate of Work Time1

first pump of first pump second pump of second pump

+ =

Step 2Well need the work rates of each pump and for that we can use the information we have in the problemstatement on each pump working individually and the following word equation for each pump doing thejob individually.

Work Rate Work Time1

of pump of pump =

For the first pump we have,

( )Work Rate 1

7 1 Work Rate of first pump =of first pump 7

=

and for the second pump we have,

( ) Work Rate 112 1 Work Rate of second pump =of second pump 12

=

Step 3Now let tbe the amount of time it takes both pumps working together to empty the pool. Using this andthe work rates we found in the second step our word equation from the first step becomes,

( )1 1

17 12

191

84

t t

t

+ =

=

Step 4

Now we can solve this for t.

19 841 4.4211

84 19t t= = =



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College Algebra

So, it will take both pumps approximately 4.4211 hours to empty the pool if they both work together.

6. John can paint a house in 28 hours. John and Dave can paint the house in 17 hours working together.How long would it take Dave to paint the house by himself?

Step 1

So, if we consider painting the house to be a single job we have the following word equation ifboth John and Dave work together to paint the house.

Portion of job Portion of job1 Job

done by John done by Dave

+ =


Work Rate Work Time Work Rate Work Time1

of John of John of Dave of Dave

+ =

Step 2We know that John can paint the house in 28 hours so we can use the following equation to determine theJohns work rate.

( )Work Rate Work Time Work Rate 1

28 1 Work Rate of John =of John of John of John 28

= =

Similarly, if we let tbe the amount of time it takes Dave to paint the house by himself we have the

following relationship between the time and work rate of Dave.

( )Work Rate Work Time Work Rate 1

1 Work Rate of Dave =of Dave of Dave of Dave

tt

= =

Step 3We can now plug in the information from the second step as well as the fact that it takes John and Dave17 hours to paint the house by themselves into the word equation from the first step to get,

( ) ( )1 1

17 17 128

17 171

28

t

t

+ =

+ =

Step 4

Now we can solve this for t.



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17 171

28

17 11

28

111728

47643.2727

11

t

t

t

t t

=

=

=

= =

So, it will take Dave approximately 43.2727 hours to paint the house by himself.

7. How much of a 20% acid solution should we add to 20 gallons of a 42% acid solution to get a 35%acid solution?

Step 1

Well start by letting xbe the amount of the 20% solution well need. This in turn means that

well have 20x + gallons of the 35% solution once were done mixing.

The basic word equation is then,

Amount of acid Amount of acid Amount of acid

in 20% solution in 42% solution in 35% solution

+ =

We know that Amount of Acid in Solution = Percentage of Solution X Volume of Solution. This gives

the following word equation.

( ) ( ) ( )Volume of Volume of Volume of

0.20 0.42 0.3520% solution 42% solution 35% solution

+ =

Step 2So, plugging all the known information in gives the following equation that we can solve for x.

( )( ) ( )0.2 0.42 20 0.35 20

0.2 8.4 0.35 7

0.15 1.4

9.33

x x

x x

x

x

+ = +

+ = +

=

=

So, well need 9.33 gallons of the 20% acid solution.

8. We need 100 liters of a 25% saline solution and we only have a 14% solution and a 60% solution.How much of each should we mix together to get the 100 liters of the 25% solution?



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Step 1

Well start by letting xbe the amount of the 14% solution well need. This in turn means that

well need 100 x gallons of the 60% solution.

The basic word equation is then,

Amount of salt Amount of salt Amount of salt

in 14% solution in 60% solution in 25% solution

+ =

We know that Amount of Salt in Solution = Percentage of Solution X Volume of Solution. This gives thefollowing word equation.

( ) ( ) ( )Volume of Volume of Volume of

0.14 0.6 0.2514% solution 60% solution 25% solution

+ =


( ) ( )0.14 0.6 100 0.25 100

0.14 60 0.6 25

0.46 35

76.09

x x

x x

x

x

+ =

+ =

=

=

So, well need 76.09 liters of the 14% saline solution and 23.91 liters of the 60% saline solution.

9. We want to fence in a field whose length is twice the width and we have 80 feet of fencing material. Ifwe use all the fencing material what would the dimensions of the field be?

Step 1

Well start by letting xbe width of the field and so 2xwill be the length of the field.

Next, we have the following word equation for the length of the fencing material.

( ) ( )2 Length of Fence 2 Width of Fence 80+ =


( ) ( )2 2 2 80

6 80

13.33

x x

x

x

+ =

=

=

So, the width of the fence will be 13.33 feet while the length will be 26.66 feet.



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College Algebra

Equations With More Than One Variable

1. Solve2

3 4E vr

=

for r.

Step 1Note that there quite a few solution paths that you can take to get the solution to this problem. For thissolution lets first distribute the 3vthough the parenthesis.

612

vE v

r=

Step 2

Next, lets clear the denominator out by multiplying both sides by r.

12 6Er vr v=

Step 3Now lets get all the terms with ron one side and the terms without ron the other side. Well also factorthe rout when were done as well. Doing this gives,

( )

12 6

12 6

Er vr v

E v r v

=

=

Step 4Finally, all we need to do is divide by both sides by the coefficient of the rto get,

6

12

vr

E v

=

Note that depending upon the path you chose for your solution you may have something slightly differentfor your answer. However, you could do some manipulation of your answer to make it look like mine (oryou could manipulate mine to make it look like yours).

2. Solve ( )6

4 17

hQ h

s= + for s.

Step 1



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College Algebra

Note that there quite a few solution paths that you can take to get the solution to this problem. For thissolution lets first clear the denominator out by multiplying both sides by 7s.

( ) ( ) ( )

( )

67 7 4 1

7

7 6 28 1

hQ s s h

s

sQ h s h

= +

= +

Step 2Now lets get all the terms with son one side and the terms without son the other side. Well also factorthe sout when were done as well. Doing this gives,

( )

( )

7 28 1 6

7 28 1 6

sQ s h h

Q h s h

=

=

Step 3

Finally, all we need to do is divide by both sides by the coefficient of the sto get,

( )6

7 28 1

hs

Q h=


3. Solve ( )6 4 17

hQ hs

= + for h.

Step 1Note that there quite a few solution paths that you can take to get the solution to this problem. For thissolution lets first clear the denominator out by multiplying both sides by 7s.

( ) ( ) ( )

( )

67 7 4 1

7

7 6 28 1

7 6 28 28

hQ s s h

s

sQ h s h

sQ h s sh

= +

= +

= +

We also distributed the 28sthrough the parenthesis in anticipation of the next step.

Step 2Now lets get all the terms with hon one side and the terms without hon the other side. Well also factorthe hout when were done as well. Doing this gives,



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College Algebra

( )

7 28 6 28

7 28 6 28

sQ s h sh

sQ s s h

=

=

Step 3Finally, all we need to do is divide by both sides by the coefficient of the hto get,

7 28

6 28

sQ sh

s

=


4. Solve

1 2 4 3

4 5

t t

A p p

+

= for t.

Step 1Note that there quite a few solution paths that you can take to get the solution to this problem. For thissolution lets first clear the denominator out by multiplying both sides by 20p.

( ) ( )

1 2 4 3

4 5

1 2 4 320 20

4 5

20 5 1 2 4 4 3

20 5 10 16 12

t tA

p p

t tp A p

p p

Ap t t

Ap t t

+ =

+ =

= +

+ = +

We also distributed the constants through the parenthesis in anticipation of the next step.

Step 2Now lets get all the terms with ton one side and the terms without ton the other side. Doing this gives,

20 21 2Ap t =

Step 3Finally, all we need to do is divide by both sides by the coefficient of the tto get,

20 21

2

Apt

=



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5. Solve10

3 7y

x=

for x.

Step 1Note that there quite a few solution paths that you can take to get the solution to this problem. For thissolution lets first clear the denominator out.

( )3 7 10

3 7 10

y x

y xy

=

=

We also distributed the y through the parenthesis in anticipation of the next step.

Step 2Now lets get all the terms withxon one side and the terms without xon the other side. Doing this gives,

7 10 3xy y =

Step 3Finally, all we need to do is divide by both sides by the coefficient of the xto get,

10 3 3 10

7 7

y yx

y y

= =

We distributed the minus sign in the denominator into the numerator to reduce the number of minus signsin the answer but doesnt need to be done if you dont want to.


6. Solve

3

12 9

x

y x

+=

for x.

Step 1Note that there quite a few solution paths that you can take to get the solution to this problem. For thissolution lets first clear the denominator out.



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College Algebra

( )12 9 3

12 9 3

y x x

y xy x

= +

= +

We also distributed the y through the parenthesis in anticipation of the next step.

Step 2Now lets get all the terms withxon one side and the terms without xon the other side. Doing this gives,

( )

12 3 9

12 3 1 9

y x xy

y x y

= +

= +

Step 3Finally, all we need to do is divide by both sides by the coefficient of the xto get,

12 3

1 9

yx

y

=

+


Quadratic Equations Part I

1. Solve the following quadratic equation by factoring.2 5 14 0u u =

Step 1Not much to this problem. We already have zero on one side of the equation, which we need to proceedwith this problem. Therefore, all we need to do is actually factor the quadratic.

( ) ( )2 7 0u u+ =

Step 2

Now all we need to do is use the zero factor property to get,

2 0 7 0OR

2 7

u u

u u

+ = =

= =

Therefore the two solutions are : 2 and 7u u= =



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College Algebra

Well leave it to you to verify that they really are solutions if youd like to by plugging them back into theequation.

2. Solve the following quadratic equation by factoring.2 15 50x x+ =

Step 1The first thing we need to do is get everything on one side of the equation and then factor the quadratic.

( )( )

2 15 50 0

5 10 0

x x

x x

+ + =

+ + =

Step 2Now all we need to do is use the zero factor property to get,

5 0 10 0OR

5 10

x x

x x

+ = + =

= =

Therefore the two solutions are : 5 and 10x x= =


3. Solve the following quadratic equation by factoring.

2 11 28y y=


( ) ( )

2 11 28 0

4 7 0

y y

y y

+ =

=


4 0 7 0

OR4 7

y y

y y

= =

= =

Therefore the two solutions are : 4 and 7y y= =




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4. Solve the following quadratic equation by factoring.219 7 6x x=


( ) ( )

26 19 7 0

3 1 2 7 0

x x

x x

+ =

+ =


3 1 0 2 7 0

OR1 7

3 2

x x

x x

= + =

= =

Therefore the two solutions are : 713 2

andx x= =


5. Solve the following quadratic equation by factoring.26 5w w =


( ) ( )

26 5 0

6 5 1 0

w w

w w

=

+ =


6 5 01 0

OR51

6

ww

ww

+ = =

==

Therefore the two solutions are : 56

and 1w w= =




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6. Solve the following quadratic equation by factoring.2 16 61 2 20z z z + =


( )

2

2

18 81 0

9 0

z z

z

+ =

=

Step 2

From the factored form we can quickly see that the solution is : 9z=

7. Solve the following quadratic equation by factoring.212 25x x=


( )

212 25 0

12 25 0

x x

x x

=

=

Make sure that you do not just cancel an xfrom both sides of the equation!


12 25 00

OR 25

12

xx

x

==

=

Therefore the two solutions are : 2512

0 andx x= =

Note that if wed canceled anxfrom both sides of the equation in the first step we would have missed the

solution 0x= !

8. Use factoring to solve the following equation.4 3 22 3 0x x x =

Step 1Do not let the fact that this equation is not a quadratic equation convince you that you cant do it! Note

that we can factor an 2x out of the equation. Doing that gives,



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( )2 2 2 3 0x x x =

The quantity in the parenthesis is a quadratic and we can factor it. The full factoring of the equation isthen,

( ) ( )2 3 1 0x x x + =


2 3 0 1 00OR OR

3 10

x xx

x xx

= + ==

= = =

Therefore the three solutions are : 0, 3 and 1x x x= = =

9. Use factoring to solve the following equation.5 39t t=

Step 1Do not let the fact that this equation is not a quadratic equation convince you that you cant do it! Note

that we move both terms to one side we can factor a 3t out of the equation. Doing that gives,

( )

5 3

3 2

9 0

9 0

t t

t t

=

=

The quantity in the parenthesis is a quadratic and we can factor it. The full factoring of the equation isthen,

( )( )3 3 3 0t t t + =


3 3 0 3 00OR OR

3 30

t tt

t tt

= + ==

= = =

Therefore the three solutions are : 0, 3 and 3t t t= = =

10. Use factoring to solve the following equation.



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2 104 3

2

ww w

w

+ =

+

Step 1This is an equation containing rational expressions so we know that the first step is to clear out the

denominator by multiplying by the LCD, which is 2w + in this case. Also, note that we now know thatwe must avoid 2w= so we do not get division by zero.

Multiplying by the LCD and doing some basic simplification gives,

( ) ( ) ( )

( ) ( ) ( ) ( )

2

2

2 2 2

2

102 4 3 2

2

10 4 2 3 2

10 2 8 6

12 0

ww w w w

w

w w w w w

w w w w w

w w

+ + = +

+

+ + = +

+ =

=

Step 2We can now factor the quadratic to get,

( )( )4 3 0w w + =

The zero factor property now tells us,

4 0 3 0OR

4 3

w w

w w

= + =

= =

Therefore the two solutions are : 4 and 3w w= = .

Note as well that because neither of these are 2w= we know that we wont get division by zero. Donot forget this important part of the solution process for equations involving rational expressions!


2

4 5 6 5

1

z z

z z z z

++ =

+ +


denominator by multiplying by the LCD, which is ( )1z z+ in this case. Also, note that we now know

that we must avoid 0z= and 1z= so we do not get division by zero.




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College Algebra

( ) ( )

( ) ( ) ( )

2

2

2

4 5 6 51 1

1

4 5 1 6 5

4 5 5 6 5

4 0

z zz z z z

z z z z

z z z z

z z z

z z

+ + + = +

+ +

+ + = +

+ + = +

=


( )4 1 0z z =


4 1 00

OR 1

4

zz

z

==

=

Note that we cannot use the first potential solution since that would give us division by zero! Therefore

the only solution is : 14

z= .

When dealing with equations that have rational expressions do not forget to verify that you do not getdivision by zero with any of the potential solutions! As we saw in this case if we had not checked wewould have gotten a value of zthat seemed to be a solution but in fact was not because of the division byzero issue.


2 7 5 81

5 5

x xx

x x

++ =

+ +


denominator by multiplying by the LCD, which is 5x + in this case. Also, note that we now know thatwe must avoid 5x= so we do not get division by zero.




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( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )2

2

2 7 5 85 1 5

5 5

2 7 5 85 1 5 5

5 5

5 1 2 7 5 86 5 2 7 5 8

9 20 0

x xx x x

x x

x xx x x x

x x

x x x xx x x x

x x

+ + + = +

+ +

+ + + = + +

+ +

+ + = ++ + =

+ + =


( ) ( )4 5 0x x+ + =


4 0 5 0OR

4 5

x x

x x

+ = + =

= =

Note that we cannot use the second potential solution since that would give us division by zero!

Therefore the only solution is : 4x= .

When dealing with equations that have rational expressions do not forget to verify that you do not getdivision by zero with any of the potential solutions! As we saw in this case if we had not checked wewould have gotten a value of xthat seemed to be a solution but in fact was not because of the division byzero issue.

13. Use the Square Root Property to solve the equation.29 16 0u =

Step 1There really isnt too much to this problem. Just recall that we need to get the variable on one side of theequation by itself with a coefficient of one. For this problem that gives,

2

2

9 16

16

9

u

u

=

=

Step 2Now all we need to do is use the Square Root Property to get,

16 16 4

9 39u= = =



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So we have the following two solutions : 4 43 3

andu u= = .

14. Use the Square Root Property to solve the equation.2 15 0x + =

Step 1There really isnt too much to this problem. Just recall that we need to get the variable on one side of theequation by itself with a coefficient of one. For this problem that gives,

2 15x =

Step 2Now all we need to do is use the Square Root Property to get,

15 15x i= =

So we have the following two solutions : 15 and 15x i x i= = .

Do not get excited about complex solutions. They will happen fairly regularly when solving quadraticequations so we need to be able to deal with them.

15. Use the Square Root Property to solve the equation.

( )2

2 36 0z =

Step 1There really isnt too much to this problem. Just recall that we need to get the squared term on one sideof the equation by itself with a coefficient of one. For this problem that gives,

( )2

2 36z =

Step 2Using the Square Root Property gives,

2 36 6z = =

To finish this off all we need to do then is solve for zby adding 2 to both sides. This gives,

2 6 2 6 4, 2 6 8z z z= = = = + =



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So, after we did a little arithmetic, have the following two solutions : 4 and 8z z= = .

16. Use the Square Root Property to solve the equation.

( )26 1 3 0t+ + =

Step 1There really isnt too much to this problem. Just recall that we need to get the squared term on one sideof the equation by itself with a coefficient of one. For this problem that gives,

( )2

6 1 3t+ =

Step 2Using the Square Root Property gives,

6 1 3 3t i+ = =

To finish this off all we need to do then is solve for tby subtracting 1 from both sides and then dividingby the 6. This gives,

6 1 3

1 3 1 3

6 6 6

t i

it i

=

= =

Note that we did a little rewrite after dividing by the 6 to put the answer in a more standard form for

complex numbers.

We then have the following two solutions : 3 31 16 6 6 6

andt i t i= = + .

Quadratic Equations Part II

1. Complete the square on the following expression.

2 8x x+

Step 1First we need to identify the number we need to add to this. Recall that we will need the coefficient of the

xto do this. The number we need is,



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College Algebra

( )2

284 16

2

= =

Step 2

To complete the square all we need to do then is add this to the expression and factor the result. Doingthis gives,

( )22 1 468x x x+ = ++


2 11u u

Step 1First we need to identify the number we need to add to this. Recall that we will need the coefficient of theuto do this. The number we need is,

( )

( )

22

2

1111 121

2 42

= =

Step 2To complete the square all we need to do then is add this to the expression and factor the result. Doingthis gives,

22 11112

121

4u u u =

+

Recall that this will always factor as uplus the number inside the parenthesis in the first step, 112

in this

case.

Do not get too excited about the fractions that can show up in these problems. They will be thereoccasionally and so we need to be able to deal with them. Luckily, if you can recall the trick to thefactoring they arent all that bad.


22 12z z

Step 1Remember that prior to completing the square we need a coefficient of one on the squared variable.However, we cant just cancel it since that requires an equation which we dont have.



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College Algebra

Therefore, we need to first factor a 2 out of the expression as follows,

( )2 22 12 2 6z z z z =

We can now proceed with completing the square on the expression inside the parenthesis.

Step 2Next well need the number we need to add onto the expression inside the parenthesis. Well need thecoefficient of the zto do this. The number we need is,

( )2

263 9

2

= =

Step 3To complete the square all we need to do then is add this to the expression inside the parenthesis andfactor the result. Doing this gives,

( ) ( )22 22 12 2 6 39 2z z z z z = = ++

Be careful when the coefficient of the squared term is not a one! In order to get the correct answer tocompleting the square we must have a coefficient of one on the squared term!

4. Solve the following quadratic equation by completing the square.

2 10 34 0t t + =

Step 1First, lets get the equation put into the form where all the variables are on one side and the number is onthe other side. Doing this gives,

210 34t t =

Step 2We can now complete the square on the expression on the left side of the equation.

The number that well need to do this is,

( )2

210 5 252

= =

Step 3At this point we need to recall that we have an equation here and what we do to one side of the equationwe also need to do the other. In other words, dont forget to add the number from the previous step toboth sides of the equation from Step 1.



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College Algebra

( )

2

2

10 325 254

5 9

t t

t

+ =

+

=

Step 4Now all we need to do to finish solving the equation is to use the Square Root Property on the equation

from the previous step. Doing this gives,

5 9 3

5 3

t i

t i

= =

=

The two solutions to this equation are then : 5 3 and 5 3t i t i= = + .


2 8 9 0v v+ =


28 9v v+ =



( )2

284 16

2

= =


( )

2

2

168 9 1

4 25

6v v

v

+ ++ =

+ =

Step 4Now all we need to do to finish solving the equation is to use the Square Root Property on the equationfrom the previous step. Doing this gives,



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College Algebra

4 25 5

4 5 4 5 9, 4 5 1

v

v v v

+ = =

= = = = + =

The two solutions to this equation are then : 9 and 1v v= = .


2 9 16 0x x+ + =


2 9 16x x+ =



( )

( )

22

2

99 81

2 42

= =

Step 3At this point we need to recall that we have an equation here and what we do to one side of the equation

we also need to do the other. In other words, dont forget to add the number from the previous step toboth sides of the equation from Step 1.

2

2

81 81

4 49 16

9 17

2 4

x x

x

+ =

+ =

+ +


9 17 17 17

2 4 24

9 17

2 2

x

x

+ = = =

=



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College Algebra

The two solutions to this equation are then :9 17 9 17

and2 2 2 2

x x= = + .

Often we will get messy answers when using completing the square to solve equations. This is notsomething to get too excited about as many applications that involve solving quadratic equations have this

kind of solution and so it is something that we just need to be able to deal with.


24 8 5 0u u + =

Step 1First, lets get the equation put into the form where all the variables are on one side, with a coefficient of

one on the 2u , and the number is on the other side. Doing this gives,

2

2

52 04

52

4

u u

u u

+ =

=



( )

222

1 12

= =


( )

2

2

52

41 1

11

4

u u

u

+

+ =

=




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College Algebra

1 1 11

4 24

11

2

u i i

u i

= = =

=

The two solutions to this equation are then :1 1

1 and 12 2

u i u i= = + .


22 5 3 0x x+ + =

Step 1

First, lets get the equation put into the form where all the variables are on one side, with a coefficient ofone on the 2x , and the number is on the other side. Doing this gives,

2

2

5 30

2 2

5 3

2 2

x x

x x

+ + =

+ =



2

25

5 2522 4 16

= =


2

2

25 255 3

2 2

5 1

4 6

16

1

16x x

x

+ =

+ =

+ +

Step 4



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College Algebra

Now all we need to do to finish solving the equation is to use the Square Root Property on the equationfrom the previous step. Doing this gives,

5 1 1

4 16 4

5 1 5 1 3 5 1, 14 4 4 4 2 4 4

x

x x x

+ = =

= = = = + =

The two solutions to this equation are then :3

and 12

x x= = .

9. Use the quadratic formula to solve the following quadratic equation.

2

6 4 0x x + =

Step 1There really isnt too much to this problem. First we need to identify the values for the quadratic formula.

1 6 4a b c= = =

Step 2Plugging these into the quadratic formula gives,

( ) ( ) ( ) ( )

( )

26 6 4 1 4 6 20 6 2 5

3 5

2 1 2 2

x

= = = =

The two solutions to this equation are then : 3 5 and 3 5x x= = + .


29 6 101w w =

Step 1First we need to get the quadratic equation in standard form. This is,

29 6 101 0w w =

Step 2Now we need to identify the values for the quadratic formula.

9 6 101a b c= = =



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College Algebra


( ) ( ) ( ) ( )

( )

( ) ( )2

6 6 4 9 101 6 36 1026 3672

2 9 18 18

6 6 102 1 102

18 3

w

= = =

= =


and3 3 3 3

w w= = + .


28 5 70 5 7u u u+ + =

Step 1First we need to get the quadratic equation in standard form. This is,

28 12 65 0u u+ + =


8 12 65a b c= = =


( ) ( ) ( )

( )

212 12 4 8 65 12 1936 12 44 3 11

2 8 16 16 4

i iu

= = = =


and4 4 4 4

u i u i

= = + .


2169 20 4 0t t + =

Step 1



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College Algebra

First we need to get the quadratic equation in standard form. This is,

24 20 169 0t t + =

Note that in this case we just rearranged the terms to have decreasing exponents.


4 20 169a b c= = =


( ) ( ) ( ) ( )

( )

220 20 4 4 169 20 2304 20 48 5

62 4 8 8 2

it i

= = = =

The two solutions to this equation are then :5 5

6 and 62 2

t i t i= = + .


2 22 72 2 58z z z z+ = +

Step 1

First we need to get the quadratic equation in standard form. This is,

2 3 130 0z z+ =

Note that in this case we just rearranged the terms to have decreasing exponents.


1 3 130a b c= = =

Step 3

Plugging these into the quadratic formula gives,

( ) ( ) ( )

( )

23 3 4 1 130 3 529 3 23

2 1 2 2z

= = =

In this case because there are no roots or complex numbers we can go further to reduce the solutions tonicer values.



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College Algebra

3 23 3 2313, 10

2 2z z

+= = = =

The two solutions to this equation are then : 13 and 10z z= = .

As a final comment we can also note that because the solutions where integers we could have also gottenthe answer by factoring! The quadratic (once written in standard form) factors as,

( ) ( )13 10 0z z+ =

and this clearly would arrive at the same solutions.

The point of all this is to note that if more than one technique can be used it wont matter which we use.Regardless of the solution technique used you will arrive at the same solutions.

Solving Quadratic Equations : A Summary

1. Use the discriminant to determine the type of roots for the following equation. Do not find any roots.

2169 182 49 0x x + =

Step 1There really isnt too much to this problem. First we need to identify the values for computing the

discriminant.

169 182 49a b c= = =

Step 2Plugging these into the formula for the discriminant gives,

( ) ( )( )22 4 182 4 169 49 0b ac = =

Step 3The discriminant is zero and so we know that this equation will have a double root.


2 28 61 0x x+ + =

Step 1



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College Algebra

There really isnt too much to this problem. First we need to identify the values for computing thediscriminant.

1 28 61a b c= = =

Step 2

Plugging these into the formula for the discriminant gives,

( ) ( ) ( )22 4 28 4 1 61 540b ac = =

Step 3The discriminant is positive and so we know that this equation will have two real roots.


249 126 102 0x x + =

Step 1There really isnt too much to this problem. First we need to identify the values for computing thediscriminant.

49 126 102a b c= = =


( ) ( ) ( )

22

4 126 4 49 102 4116b ac = =

Step 3The discriminant is negative and so we know that this equation will have two complex roots.


29 151 0x + =

Step 1There really isnt too much to this problem. First we need to identify the values for computing thediscriminant.

9 0 151a b c= = =




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College Algebra

( ) ( ) ( )22 4 0 4 9 151 5436b ac = =

Step 3The discriminant is negative and so we know that this equation will have two complex roots.

Application of Quadratic Equations

1. The width of a rectangle is 1 m less than twice the length. If the area of the rectangle is 100 m2whatare the dimensions of the rectangle?

Step 1

Well start by letting Lbe the length of the rectangle. From the problem statement we now know that thewidth of the rectangle is 1 m less than twice the length and so must be 2 1L .

Step 2

We also know that the area of any rectangle is length times width and we are given that the areaof this particular rectangle is 100. Therefore the equation for this problem is,

( ) ( )

( )( )2

length width

100 2 1

100 2

A

L L

L L

=

=

=

Step 3This is a quadratic equation and we know how to solve that so lets do that. First, we need to get thequadratic equation in standard form.

22 100 0L L =

We can now use the quadratic formula on this to get,

( ) ( ) ( )( )

( )

21 1 4 2 100 1 801

2 2 4L

= =

Step 4Reducing the two values we got in the previous steps to decimals we arrive at the following two solutionsto the quadratic equation from Step 2.

1 801 1 8016.8255 7.3255

4 4L L

+= = = =



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College Algebra

Step 4Next, we know that we can find the distance of each car using the formula,

( )( )Distance Speed of Car Time driving=

So, for each car we have,

( )( )

( ) ( ) ( )

Distance of Car A 40 40

Distance of Car B 60 3 60 3

t t

t t

= =

= =

Putting all of this into the word equation we wrote down in Step 3 we get the following equation.

( ) ( )( )

( )

( )

22

22 2 2

2 2

2 2

2

40 60 3 250,000

40 60 3 250,000

1600 3600 6 9 250,000

1600 3600 21,600 32, 400 250,000

5200 21,600 217,600 0

t t

t t

t t t

t t t

t t

+ =

+ =

+ + =

+ + =

=

Note as well that we did quite a bit of simplification to get the equation into a standard form. Also, do notget excited about the large numbers here! They happen on occasion so they are nothing to worry about.This is still just a quadratic and we know how to solve quadratic equations. It doesnt matter if thenumbers are single digit numbers of significantly larger numbers as they are here.

Step 5As noted in the previous step this is just a quadratic equation and we know how to solve those!Using the quadratic formula gives,

( ) ( ) ( ) ( )

( )

221,600 21,600 4 5200 217,600 21,600 4,992,640,000

2 5200 10,400t

= =


21,600 4,992,640,000 21,600 4,992,640,0004.7172 8.8710

10,400 10,400t t

+= = = =

The first solution to the equation doesnt make any sense since it is negative (we are working with time

and so its safe to assume we are starting at 0t= after all!) so that means the second is the answer weneed.



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This means that Car A (i.e.the one traveling at 40 mph) travels for 8.871 hours while Car B (i.e.the onetraveling at 60 mph) travels for 5.871 hours (three hours less than Car A time!).

3. Two people can paint a house in 14 hours. Working individually one of the people takes 2 hours more

than it takes the other person to paint the house. How long would it take each person workingindividually to paint the house?

Step 1First, let Person A be the faster of the two painters and let tbe the amount of time it takes to paint the

house by himself. Next, let Person B be the slower of the two painters and so it will take this person 2t+ hours to paint the house by himself.

Step 2

Working together they can paint the house in 14 hours so we have the following word equationfor them working together to paint the house.

Portion of job Portion of job1 Job

done by Person A done by Person B

+ =


( ) ( )

Work Rate Work Time Work Rate Work Time1

of Person A of Person A of Person B of Person B

Work Rate Work Rate14 14 1

of Person A of Person B

+ =

+ =

Step 3Now we need the work rate of each person which we can get from their individual painting times asfollows,

( )Work Rate Work Time Work Rate

1of Person A of Person A of Person A

1Work Rate of Person A =

t

t

= =

( )

Work Rate Work Time Work Rate

2 1of Person B of Person B of Person B

1Work Rate of Person B =

2

t

t

= + =

+

Step 4Plugging these into the word equation from Step 2 we arrive at the following equation.



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College Algebra

( ) ( )1 1

14 14 12

14 141

2

t t

t t

+ =

+

+ =+

Step 5

To solve this we know that well need to multiply by the LCD, ( )2t t+ in this case, to clear thedenominators. Doing this gives,

( ) ( )2

2

14 2 14 2

28 28 2

26 28 0

t t t t

t t t

t t

+ + = +

+ = +

=

After some simplification we arrive a fairly simple quadratic equation to solve. Using the quadraticformula gives,

( ) ( ) ( )( )

( )

226 26 4 1 28 26 788

2 1 2L

= =


26 788 26 7881.0357 27.0357

2 2t t

+= = = =

The first solution to the equation doesnt make any sense since it is negative (we are working with time

and so its safe to assume we are starting at 0t= after all!) so that means the second is the answer weneed.

This means that Person A can paint the house in 27.0357 hours while Person B can paint the house in29.0357 hours (two hours more than Person A).

Equations Reducible to Quadratic Form

1. Solve the following equation.6 39 8 0x x + =



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College Algebra

Hint : Remember to look at the exponents of the first two terms and try to find a substitution that will turnthis into a normal quadratic equation.

Step 1

First lets notice that ( )6 2 3= and so we can use the following substitution to reduce the equation to a

quadratic equation.

( )2

3 2 3 6u x u x x= = =

Step 2Using this substitution the equation becomes,

( ) ( )

2 9 8 0

1 8 0

u u

u u

+ =

=

We can easily see that the solution to this equation is : 1u=

and 8u=

.

Step 3Now all we need to do is use our substitution from the first step to determine the solution to the originalequation.

( )1

331: 1 1 1u x x= = = =

( )1

338 : 8 8 2u x x= = = =

Therefore the two solutions to the original equation are : 1 and 2x x= = .

2. Solve the following equation.4 27 18 0x x =


Step 1

First lets notice that ( )4 2 2 = and so we can use the following substitution to reduce the equation toa quadratic equation.

( )2

2 2 2 4u x u x x

= = =




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College Algebra

( ) ( )

2 7 18 0

9 2 0

u u

u u

=

+ =

We can easily see that the solution to this equation is : 2u= and 9u= .


2 2

2

1 1 1 12 : 2

2 2 2u x x x i

x

= = = = = =

2 2

2

1 1 1 19 : 9

9 9 3u x x x

x

= = = = = =

Therefore the four solutions to the original equation are :

1 1

and 32x i x= =

.

3. Solve the following equation.2 1

3 34 21 27 0x x+ + =


Step 1First lets notice that ( )2 13 32= and so we can use the following substitution to reduce the equation to aquadratic equation.

21 1 2

23 3 3u x u x x

= = =


( ) ( )

2

4 21 27 04 9 3 0u uu u

+ + =+ + =

We can easily see that the solution to this equation is : 94

u= and 3u= .




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College Algebra

31

39 9 9 729

:4 4 4 64

u x x

= = = =

( )1

333: 3 3 27u x x= = = =

Therefore the two solutions to the original equation are :729

and 2764

x x= = .

4. Solve the following equation.8 46 7 0x x + =

Hint : Remember to look at the exponents of the first two terms and try to find a substitution that will turn

this into a normal quadratic equation.

Step 1

First lets notice that ( )8 2 4= and so we can use the following substitution to reduce the equation to aquadratic equation.

( )2

4 2 4 8u x u x x= = =


2 6 7 0u u + =

Now, this equation does not factor. That happens on occasion but luckily enough we know how to solveit anyway. All we need to do is use the quadratic formula to find the solutions to this equation.

( ) ( ) ( ) ( )

( )

26 6 4 1 7 6 8 6 2 2

3 22 1 2 2

u

= = = =

Do not get excited about the messy numbers here! These kinds of solutions happen on occasion and wejust need to be able to deal with them. Just keep in mind that they are just numbers even if they are not

the integers we are used to seeing!


( ) ( )1 1

4 4 43 2 : 3 2 3 2 4.4142 1.4495u x x= + = + = + = =



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( ) ( )1 1

4 4 43 2 : 3 2 3 2 1.5858 1.1222u x x= = = = =

Therefore the two solutions to the original equation are : 1.222 and 1.4495x x= = .

5. Solve the following equation.

2

2 1721 0

x x+ + =

Hint : This works exactly the same as the first four problems even though thexs are in the denominator.The only difference here is that the xs will be in the denominator of our substitution.

Step 1

First lets notice that ( )2 2 1= and so we can use the following substitution to reduce the equation to aquadratic equation.

2 22

2 2

1 1 1u u

x x x x

1 = = = =


( ) ( )

22 17 21 0

2 3 7 0

u u

u u

+ + =

+ + =

We can easily see that the solution to this equation is : 32

u= and 7u= .


3 1 3 1 2:

32 2 3

2

u xx

= = = =

1 1 17 : 7 7 7u xx= = = =

Therefore the two solutions to the original equation are : 2 13 7

andx x= = .




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1 1118 0

x x + =

Hint : This works exactly the same as the first four problems even though thexs are in the denominator.The only difference here is that the xs will be in the denominator of our substitution.

Step 1

First lets notice that ( )121 2= and recall that1

2x x= . So we can use the following substitution to

reduce the equation to a quadratic equation.

22

2

1

2

1 1 1u u

xx xx

1 = = = =

Step 2

Using this substitution the equation becomes,

( ) ( )

2 11 18 0

2 9 0

u u

u u

+ =

=

We can easily see that the solution to this equation is : 2u= and 9u= .


2

1 1 1 12 : 22 2 4

u x xx

= = = = =

21 1 1 1

9 : 99 9 81

u x xx

= = = = =

Therefore the two solutions to the original equation are : 1 14 81

andx x= = .

Equations with Radicals


2 3x x= +



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Step 1The first step here is to square both sides to get,

( ) ( )22

2

2

2 3

4 34 3 0

x x

x xx x

= +

= +

=

Step 2This is just a quadratic equation and we know how to solve it so lets do that.

( ) ( )3

4 3 1 0 , 14

x x x x+ = = =

As shown we have two solutions to the quadratic we got from the first step.

Hint : Recall that the solution process used here can, and often does, introduce values that are not in factsolutions to the original equation!

Step 3Were not done with this problem. Recall from the notes that the solution process we used here has theunfortunate side effect of sometimes introducing values that are not solutions to the original equation.

So, to finish this out we need to check both of the potential solutions from the previous step in the originalequation (recall its important to check the potential solutions in the original equation).

( )? ?

3 3 3 9 3 34 4 2 4 2 2

3: 2 3 NOT OK

4x= = + =

( ) ? ?1: 2 1 1 3 2 4 2 2 OK x= = + = =

Only one of the potential solutions work out and so the original equation has a single solution : 1x= .


33 2 1x x = +

Step 1

The first step here is to square both sides to get,

( ) ( )2 2

2

2

33 2 1

33 2 2 1

4 32 0

x x

x x x

x x

= +

= + +

+ =

Step 2



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This is just a quadratic equation and we know how to solve it so lets do that.

( ) ( )8 4 0 8, 4x x x x+ = = =



Step 3Were not done with this problem. Recall from the notes that the solution process we used here has theunfortunate side effect of sometimes introducing values that are not solutions to the original equation.


( )? ?

8: 33 2 8 8 1 49 7 7 7 NOT OK x= = + =

( )? ?

4 : 33 2 4 4 1 25 5 5 5 OK x= = + = =



7 39 3x x= +

Step 1The first step here is to square both sides. However, before we do that we need to get the root on one sideby itself.

7 39 3x x+ = +

Now we can square both sides to get,

( ) ( )22

2

2

7 39 3

14 49 39 3

11 10 0

x x

x x x

x x

+ = +

+ + = +

+ + =


( ) ( )10 1 0 10, 1x x x x+ + = = =




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Step 3Were not done with this problem. Recall from the notes that the solution process we used here has the

unfortunate side effect of sometimes introducing values that are not solutions to the original equation.


( ) ( )? ?

10 : 7 39 3 10 10 7 9 10 7 13 NOT OK x= = + = +

( ) ( )? ?

1: 7 39 3 1 1 7 36 1 7 7 OK x= = + = + =



1 2 2x x= +

Step 1The first step here is to square both sides. However, before we do that we need to get the root on one sideby itself.

1 2 2x x =

Now we can square both sides to get,

( ) ( )22

2

2

1 2 2

2 1 2 2

4 3 0

x x

x x x

x x

=

+ =

+ =


( ) ( )1 3 0 1, 3x x x x = = =



Step 3



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Were not done with this problem. Recall from the notes that the solution process we used here has theunfortunate side effect of sometimes introducing values that are not solutions to the original equation.


( )? ?

1: 1 1 2 1 2 1 1 0 1 1 OK x= = + = + =

( )? ?

3: 3 1 2 3 2 3 1 4 3 3 OK x= = + = + =

Both of the potential solutions work out and so the original equation has a two solutions :

1 and 3x x= = .


1 1 2 4x x+ = +

Step 1The first step here is to square both sides to get,

( ) ( )

( )

2 2

2

1 1 2 4

1 2 1 1 2 4

1 2 1 1 2 4

2 1 3 2

x x

x x x

x x x

x x

+ = +

+ + = +

+ + = +

= +

Step 2Unlike the previous problems squaring both sides once isnt sufficient to eliminate the square roots fromthe problem. So, once we get the remaining root on one side by itself, as we did in the previous step, weneed to square both sides once again.

Doing that gives,

( ) ( )

( )

2 2

2

2

2 1 3 2

4 1 9 12 4

9 16 0

x x

x x x

x x

= +

= + +

+ =


( )16

9 16 0 0,9

x x x x+ = = =



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3 30z >

Note that we could just have easily subtracted 4zfrom both sides and added 23 to both sides. Each willget the same result in the end.

Step 3

For the final step we need to divide both sides by -3. Recall however that because we are dividing by anegative number we need to switch the direction of the inequality to get,

10z<

So, the inequality form of the solution is 10z< and the interval notation form of the solution is

( ),10 .

Remember that we use a parenthesis, i.e.), for the right side of the interval notation because we are notincluding 10 in the solution. Also recall that infinities always get parenthesis!

2. Solve the following inequality and give the solution in both inequality and interval notation.

( ) ( )1 1 1 1

3 4 6 2 102 3 2 4

t t t

+ +

Hint : Remember that solving linear inequalities is pretty much the same as solving a linear equation. Justremember to be careful when multiplying/dividing by a negative number.

Step 1We know that the process of solving a linear inequality is pretty much the same process as solving a

linear equation. We do basic algebraic manipulations to get all the ts on one side of the inequality andthe numbers on the other side. Just remember that what you do to one side of the inequality you have todo to the other side as well. So, lets go through the solution process for this linear inequality.

First, we should clear out the parenthesis on both sides and do any simplification that we can. Doing thisgives,

3 1 52 2 3

2 2 2

3 3 112

2 2 2

t t t

t t

+

+

Step 2

We can now add 112t to both sides and subtract 3

2from both sides to get,

150

2t


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Alg Solving Solutions

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