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harpazo1965 wrote:Complete the square.
ax^2 + ba + c = 0, where a > 1.
If the middle term is "ba", then there is no middle term, so the square-completion process is nothing more than isolating the one square:
Since anything can be viewed as having a zero added to it, then:
If the middle term is meant to be "bax", then try following the steps shown here: https://www.mathisfunforum.com/viewtopic.php?pid=435501#p435501
Please, read what I said to Bob. This is not my own created problem. Understand? I found it online.
ax^2 + ba + c = 0, where a > 1.
Did you mean ax^2 + bx + c = 0, where a > 1.
B
It's not what I mean. I found the problem online. It is not my own created question.
harpazo1965 wrote:What is the best way to solve tricky or lengthy word problems?
Take your time, and be clear on your reasoning. Otherwise, it's just a matter of practice, practice, practice.
1. It requires more than practice.
2. I think it also requires learning clear, to the point methods.
3. Practice without understanding leads to zero learning.
harpazo1965 wrote:Why does the radicand come out?
See here:
sqrt{(a + b)^2} = (a + b)
Why is that the case?
Squaring and square-rooting are inverse operations (assuming the argument of the square root is non-negative). They undo each other. It is the same reason that 1 + 1 - 1 = 1 or (2*3)/3 = 2.
Yes, this is what Bob said. By the way, I left the other math forum after I was so rudely insulted by Jonah. Do not tell him that I am here. If you do, I will request for you to be banned from this site. He does not need to know where I am. Agree?
Square and square root are inverse operations. If you do one followed by the other you get back to where you started. Like double and half
B
It makes sense, Bob. Ok. Cool.
Yes.
But try 2^3^5
B
Let me see.
3^5 = 243
So, we now have 2^(243). I think this leads to a number that must be
expressed in scientific notation. Yes?
You say?
How do you solve a square root raised to another square root?
What is sqrt{2}^(sqrt{2})?
Let me see.
Should I rewrite sqrt{2} as x^(1/2)?
[x^(1/2)]^(x^(1/2))
(1/2)(x^(1/2) = (x^(1/2))/(2)
I now get x^[(x^(1/2))/(2)].
You say?
What is the best way to solve tricky or lengthy word problems?
Why does the radicand come out?
See here:
sqrt{(a + b)^2} = (a + b)
Why?
I can do it with anything.
Look here:
sqrt{(apple)^2} = (apple).
Why is that the case?
How is this done?
(x^2)^(2)^(2)
Let me see.
(x^2)^4 = x^8
Yes?
Complete the square.
ax^2 + ba + c = 0, where a > 1.
Let me see.
Divide every term by a.
This leads to x^2 + b + (c/a) = 0
Divide middle term by 2 and then square.
So, b becomes (c/2)^2 = c^2/4.
Add (c^2/4) to right side.
x^2 + b + (c^2/4) = (c^2/4)
Factor left side.
(x + (c/2)) (x + (c/2)) = (c^2/4)
(x + (c/2))^2 = (c^2/4)
Take square root on both sides.
sqrt{(x + (c/2))^2} = sqrt{(c^2/4)}
x + (c/2) = c/2
x = c/2 - c/2
x = 0
Does this make sense? Did I make an error?
Just a minute. This doesn't seem remarkable. If you were born in the year Y and your age is A
then Y + A = the current year.So next year you will be A + 1 and Y + A + 1 = 2024. Why do we have to wait 1000 years for this?
In the year of your birth your age is 0.
So Y + A = that year.
As each year passes A goes up by 1 and so does the year.
Bob
Good point, Bob. I like it.
In post #6 I have derived an equation for working this out. Do you understand it?
Bob
Yes, Bob I understand how to work it out. I will show my work on days off. I work 32 weekend hours (two back to back double shifts). I am busy and seriously tired.
Today is a day that happens every 1000 years.
If you add the year if your birth and your age it will
yield the number 2023. This I true concerning every person in the world.
Note: This is the whole problem. I don't see a question in this problem. I don't get it. How about you?
It does look that way. If you put those numbers into this:
https://www.mathsisfun.com/quadratic-eq … olver.html
The graph doesn't cross the x axis which would confirm no real roots.
Bob
Ok. No solution over the real numbers. Cool. My concern is to complete the square correctly. Thanks again.
Note:
Bob,
My family and friends are confused about love for mathematics. They say that solving textbook questions at 58 is a clear indication that something may be wrong with me. Really? What can I do at 58? I also play classical guitar hymns. They are also confused about my love for music and playing guitar.
At 58:
A. I am not popular with women anymore
B. Bald
C. Terribly overweight
D. Work For a dead-end security company
Do you ever come across people who have nothing else to do but criticize something you love to do? If math brings me joy, what's wrong with that? If playing guitar brings me joy, what's wrong with that?
You say?
This is my method:
Old volume = 12 times 7 times 3
New volume = (12-x)(7-x) times 3
10% reduction means new volume = 0.9 of old volume. So my equation isI simplified that to a quadratic, found x and checked it led to a 10% reduction. There was a second value from the quadratic which was obviously too big (ie > 12)
Bob
Wow! You are definitely talented. Trust me, there are thousands of college students and graduate students who have no idea how to even begin solving this problem. I get confused with the bad wording in most word problems. Thanks for the equation. I can take it from here.
hi
This problem is made more complicated because it switches from yards, to feet to inches. To get a correct equation everything needs to be in the same units. So I'll convert to feet.
Radius of pool = 5 ft
Let the width of the border (what we want to find) be x ft. Then radius out to the outer edge is 5 + xArea of the circle out to the outer edge is
So area of the concrete ring is
The concrete is 3 inch thick = 1/4 ft. The volume of concrete is 1 cu yard = 3 x 3 x 3 = 27 cu ft.
Volume of the border concrete = area x thickness hence
Bob
Wow! Wish I could break down the problem like you just did. I did get parts of it
correct but guessing, of course. This has been my problem with mathematics for years. I know to let x be what we are looking for but forming the needed equation(s) to solve applications has been a struggle for me. I can take it from here. Thanks for the equation.
Hi amnkb;
amnkb wrote:sry i dont know how to help on an iphone
(if youd been on an android phone then youd know what google play store is and how to search for stuff in it)
(my bad- i'd just assumed)harpazo1965 posted:
...I have an android AS 21 T-Mobile phone...
...my android phone A21...Googling "Android AS 21" failed to find one of those, but turned up these:
Samsung Galaxy A21 (released April 2020)
Samsung Galaxy A21s (released May 2020)
Samsung Galaxy S21 (released January 2021)So I think he has an Android...a Samsung Galaxy (model unclear).
I will provide the link for anyone who is interested to see the photo. My Android AS-21 does not allow picture uploads here. If there is a way to do it , I have no idea how it is done. I am moving beyond this chat.
harpazo1965 wrote:Outline: 1. [The area A of a circle with radius r is given by] A = pi•r^2
yes
harpazo1965 wrote:2. The expression that gives the outer radius is not too clear to me.
I think it is: A = pi•(2x + 10)^2draw the circle for the pool
label the radius
draw the outer cricle for the border
label the width
label the radius
is the radius of the pool+border area equal to the whole diameter of the pool plus the border width on both sides?harpazo1965 wrote:3. A foot is 12 inches. How much of a foot is 3 inches? It is 3/12 of a foot.
3/12 = 1/4 = 0.25
harpazo1965 wrote:4. Let A_b = area of border. I think it is pi(2x + 10)^2 - pi•(5)^2
no
harpazo1965 wrote:5. Is the right equation the following?
27 = pi(2x + 10)^2 - pi•(5)^2, where x is the width of the border.
where does this use the thickness of the border?
Radius of pool = 5 feet
Border area = pi(x + 5)^2
Radius of pool + border area = 2x + 10 feet.
So, 5 + pi(x + 5)^2 = 2x + 10
Correct equation?
harpazo1965 wrote:A jumbo chocolate bar with rectangular shape measures 12 cm in length, 7 cm in width, and 3 cm in thickness. Due to escalating costs of cocoa, management decides to reduce the volume of the bar by 10%. To accomplish this reduction, management decides that the new bar should have the same 3 cm thickness, but the length and width of each should be reduced an equal number of centimeters. What should be the dimensions of the new candy bar?
amnkb wrote:old length: 12
old width: 7
old thickness: 3
old volume: ___new length: 12 - x
new width: 7 - x
new thickness: 3 (same as old thickness)(sry abt typo before)
new volume: ___new volume is 90% of old volume
harpazo1965 wrote:Old volume = 7•12•3 or 252 cm^3
New volume is 2(12 - x)(7 - 2)
this assumes that x=2
how did you get this?
when you checked your ans did it work out right?harpazo1965 wrote:A. Where did 12 come from?
B. Where did 7 come from?
A jumbo chocolate bar with rectangular shape measures 12 cm in length, 7 cm in width,, and 3 cm in thickness.
harpazo1965 wrote:C. What must I do with 90%?
new volume is 90% of old volume
what is 90% of old volume?
set equal and solve
1. I made a typo, obviously.
I meant to say that the new volume is 2(12 - x)(7 -x)
2. The old volume is 252 cm^3. So, 0.90 (252) = 226.8 cm^3.
3. The set up is here:
226.8 = 2(12 - x)(7 - x)
I get two answers for x:
x_1 is approximately-20.67035
We are talking about measurement. So, the value of x_1 must be rejected.
I also know that x_2 is approximately 39.67035. I can round this to be 40.
Dimensions of the new candy bar:
Length = (12 - 40) leads to a negative answer.
Width = (7 - 40) leads to a negative answer.
Height = 2
Something is wrong. The length, width and height must be positive.
What did I do wrong?
Is 226.8 = 2(12 - x)(7 - x) the correct equation needed to find the new length, width and height of the candy bar?
amnkb wrote:open google play store
select 'search'
enter 'carbon browser'
select and installharpazo1965 wrote:This is too confusing.
amnkb wrote:which part? you installed carbon browser and then??
harpazo1965 wrote:I am not a technical person. What is a carbon browser?
sry i dont know how to help on an iphone
(if youd been on an android phone then youd know what google play store is and how to search for stuff in it)
(my bad- i'd just assumed)
It's ok. I really don't know what the big deal is with members here. Simply copy and paste the link provided in my threads to see the photo. What's wrong with that?
amnkb wrote:the complete-the-squarr process is:
half of b/a is b/2a
square this to get the completed-square 3erd term
add this squared term to both sides
including any multiplier 'a':half of (pi)/2 is (pi)/4
square this to get ((pi)^2)/16
unless they tell you specifically to round to 4 decimal places you should do exact using piharpazo1965 wrote:x^2 + (π/2) x + π^2/16 = -1 + π^2/16
(x + (π/4))(x + (π/4)) = 1.4674
like bob said: stop rounding
use exact valuesharpazo1965 wrote:(x + (π/4))^2 = -0.38315
where did -0.38315 come from?
all you did was change (expression)(expression) to (expression)^2
thats just a change of formatting
the value of the other side should not have changedharpazo1965 wrote:Stuck here....
follow the steps exactly as listed
including adding the (half of (middle coefficient))^2 to both sides
Solve equation by completing the Square.
2x^2 + π x + 2 = 0
Let me see.
In order to complete the square, the coefficient of the x square term
cannot be greater than or less than 1.
So, I must divide each term on both sides of the equation, in this case, by 2.
(2x^2 + π x + 2)/2 = 0/2
Doing so, I get the following:
x^2 + (π/2) x + 1 = 0
I now subtract 1 from both sides.
x^2 + (π/2) x = -1
The next step is to divide the coefficient of x by 2 and then raise to the second power and add the product to both sides of the equation.
[(π/2)^2]/2 = (π/4)^2 = (π^2/16)
I must now add the product to both sides of the equation.
x^2 + (π/2) x + π^2/16 = -1 + π^2/16
x^2 + (π/2) x + π/4 = -1 + π^2/16
(x + π/4)(x + π/4) = negative answer. Do you understand why I needed to subtract 1 from both sides of the equation a few steps before? Do you know where -1 comes from? I showed where -1 comes from a few steps before. Now, with the right side being negative, can we conclude that this equation has no solution over the real numbers?
You say?
harpazo1965 wrote:A circular pool measures 10 feet across. One cubic yard of concrete is to be used to create a circular border of uniform width around the pool. If the border is to have a depth of 3 inches, how wide will the border be?
I know that 1 cubic yard is 27 cubic feet.
I say let x = width of the border around the pool
I don't know where to go with this information.what is the formula for the area of a circle with radius r?
if x is the border width then what expression gives the outer radius of the border?
how much of a foot is 3 inches?
the area of the border is (outer circle - inner circle)
the volume is area times thickness
plug info into equation, set equal to 27
solve for x
Outline:
1. A = pi•r^2
2. The expression that gives the outer radius is not too clear to me.
I think it is: A = pi•(2x + 10)^2
3. A foot is 12 inches. How much of a foot is 3 inches? It is 3/12 of a foot.
4. Let A_b = area of border. I think it is pi(2x + 10)^2 - pi•(5)^2
5. Is the right equation the following?
27 = pi(2x + 10)^2 - pi•(5)^2, where x is the width of the border.
Yes? If not, can you please set up the correct equation? I can then take it from there.
Thank you.
harpazo1965 wrote:A jumbo chocolate bar with rectangular shape measures 12 cm in length, 7 cm in width, and 3 cm in thickness. Due to escalating costs of cocoa, management decides to reduce the volume of the bar by 10%. To accomplish this reduction, management decides that the new bar should have the same 3 cm thickness, but the length and width of each should be reduced an equal number of centimeters. What should be the dimensions of the new candy bar?
This one through me into a loop.
I think this problem demands the use of the volume formula
V = length • width • height. The part about reducing the volume by 10% is confusing to me.Can someone explain what is going on in this problem and set up the right equation for me to use?
old length: 12
old width: 7
old thickness: 3
old volume: ___new length: 12 - x
new width: 7 - x
new thickness: 2
new volume: ___new volume is 90% of old volume
Old volume = 7•12•3 or 252 cm^3
New volume is 2(12 - x)(7 - 2)
A. Where did 12 come from?
B. Where did 7 come from?
C. What must I do with 90%?
amnkb wrote:the complete-the-squarr process is:
half of b/a is b/2a
square this to get the completed-square 3erd term
add this squared term to both sides
including any multiplier 'a':half of (pi)/2 is (pi)/4
square this to get ((pi)^2)/16
unless they tell you specifically to round to 4 decimal places you should do exact using piharpazo1965 wrote:x^2 + (π/2) x + π^2/16 = -1 + π^2/16
(x + (π/4))(x + (π/4)) = 1.4674
like bob said: stop rounding
use exact valuesharpazo1965 wrote:(x + (π/4))^2 = -0.38315
where did -0.38315 come from?
all you did was change (expression)(expression) to (expression)^2
thats just a change of formatting
the value of the other side should not have changedharpazo1965 wrote:Stuck here....
follow the steps exactly as listed
including adding the (half of (middle coefficient))^2 to both sides
Ok. I will work it out completely later and show my work here.