Posted on

Algebra Ch13 Test A Problem #13

Question from Grace:

We are using this test A as a practice test. We can’t figure out how to solve this.

Answer from Dr. Callahan:

Grace,

To solve this one, you take the square root of each side

+/- SQRT ( 16x^2 + 8x +1 ) = +/- SQRT (9)

The left side needs a little work to do this. Best to see if we can factor
16x^2 + 8x +1

As a hint, you know your factors would need to be the same since you are taking a square root.

So start with the first term 16x^2 – square root of it is 4x. So try that

(4x + ___) (4x + ___)

Then the last term is 1. So the obvious thing that can be multiplied by itself to get 1 is 1. So try

(4x + 1) (4x + 1)

Now when I multiply is out I get

16x^2 + 4x + 4x + 1 = 16x^2 + 8x +1

So that works!!

Now back to the problem

+/- SQRT ( 16x^2 + 8x +1 ) = +/- SQRT (9)

+/- SQRT ((4x + 1) (4x + 1)) = +/- SQRT (9)

Using the + we have
(4x + 1) = 3
And 4x = 2 so x = 1/2 is one answer

Using the – we have
(4x + 1) = -3
4x+1 = -3
4x = -4
x = -1

 

 

Posted on

Algebra Ch10.2 #3

Question from Makenna

I don’t understand how the perimeter can be 678 ft.
255 is the width x 2 = 510
505 is the length x 2 = 1010 + 510 = 1520

 

Answer from Dr. Callahan

Makenna,

You are correct. The 678 was from the older version of the book (which used a flag of 104 x 235. Flags have grown since then. 😉)

Just FYI – the older flag was flown in Detroit on the side of the J.L. Hudson Store – which no longer exists.
https://en.wikipedia.org/wiki/J._L._Hudson_Department_Store_and_Addition

Posted on

Algebra Ch5.6 #15

Ch5.6 #15

As before – all are similar so I will pick the harder one.

Givens

AB = CD

AB = 9(x-2)

CD = x + 3(x+5)

Therefore,

9(x-2) = x + 3(x+5)

Now multiply out and solve

9x – 18 = x + 3x + 15

9x – x – 3x  = 15 + 18

5x = 33

x = 6.6

Now go back and find the lengths

AB = 9(x-2) = 9(6.6 – 2) = 41.4

CD = AB = 41.4

CE = x = 6.6

ED = 3(x+5) = 3(6.6+5) = 34.8

Posted on

Algebra Ch5.6 #14

Now let’s look at Ch 5.6 #14.

Parts a-c are similar – so will do part c since the harder.

Area = x + 20 = (x-1)4    > I am just stating all things that are equal based on the given.

x+20 = (x-1)4

x+20 = 4x -4  > multiplied out

20 + 4 = 4x – x  > moved similar items to same side

24 =  3x

x = 8

Now we know x, we have to find the sides (ALWAYS GO BACK AND MAKE SURE YOU KNOW WHAT WAS ASKED)

x -1  = 8 – 1 = 7

4 = 4

 

Posted on

Algebra Ch5.6 #9

For problem 5.6, parts a-d are all pretty much the same. Here I will do part d)  since a hair more complex. The steps the same for the others.

Since AB = CD, we can substitute the following

AB is x + 6(x+2)

CD is 2(x+10)

To get

x + 6(x+2)  = 2(x+10)

So to find the length of AB and CD, I need to figure out x. So solve the equation.

x + 6(x+2)  = 2(x+10)

x + 6x + 12 = 2x + 20   – I just multiplied everything out

x + 6x – 2x = 20 – 12  > I moved all the x terms to the left and all the constants (numbers) to the right

5x = 8  > did the math

x = 8/5  = 1.6

Now I have x, but the problem asked for the lengths.

So now we can find the parts

AB = x + 6(x+2) = 1.6 + 6(1.6 + 2) = 1.6 + 6(3.6) = 23.2

AE = x = 1.6

EB = 6(x+2) = 21.6

CD = AB = 23.2

 

 

To do the others – follow the same process

Posted on

Algebra Ch5.1 #10x

This is an error in the Algebra Solutions Manual

Mary asked,

In Algebra 1, chapter 5 lesson 1 problem 10x, the answer key says the answer is true for all positive integers.  But isn’t it true for all integers? 1 to the power of x =1.  If it is a negative power, it is still one, right?

Mary
Dr. Callahan Answer:
You are correct. Should be for all real numbers – not just integers. I tend to test these in a calculator to make sure though 😉
So that is incorrect in the Solutions manual.
Just for completeness sake – my goto math tool is Wolfram – so here is there answer.
 
Posted on

Algebra Textbook Evaluations

Look for the following material or heading in the contents.

  • Variables
  • Exponents
  • Order of Operations
  • Equations and inequalities
  • Word problems (converting words into symbols)
  • Real Numbers
  • Adding
  • Subtracting
  • Multiplication
  • Division
  • Distributive property
  • Linear Equations (might be equations in one variable)
  • Graphing
  • Slope
  • Intercepts
  • Point-slope and or slope-intercept formulas
  • Systems of linear equations (or simultaneous equations)
  • Linear inequalities
  • Solving
  • Graphing
  • Absolute Values
  • Exponents
  • Products
  • Divisions
  • Scientific notation
  • Polynomials
  • Quadratic equations
  • Factoring
  • Radicals

Notes for use: All the material on the above list needs to be covered in algebra. Instead of moving fast, the students should understand these concepts pretty well.

Posted on

Algebra Ch3.4 Set I #9 Solutions Manual Error

Angela’s son brought this error in the Earth cover Algebra I Solutions Manual to our attention. Thank you!

The problem (Ch3.4 #9) wording changed completely from the salamander cover to the Earth cover. However, the solution in the old and new solutions manuals is the same.  In other words, both manuals have the solution to the salamander wording of the problem. The Earth solutions manual has the wrong solution to the Earth Ch3.4 #9 problem.

Angela’s son worked it perfectly and here is his correct solution to the Earth cover Algebra I Chapter 3.4 Set 1 #9 problem.

Posted on

Algebra Ch1.Review Set I #14b Error

A question brought to us by Marybeth. Thank you!

In Jacob’s Elementary Algebra Chapter 1 Review Set I Question 14b page 51.

“What is the total number of atoms in x propane molecules as a sum?”

The Solutions Manual for Elementary Algebra on page 14 lists the answer as:

b) 3x + 5x

The “5” is a typo. The answer should be 3x + 8x.

3x carbon atoms plus 8x hydrogen atoms