This guy is the bomb!! He has such an easy, affable delivery that I think I could have learned to love math had he been my mentor 'back in the day'. I understood the mechanics back then, but never the 'why?' of what I was doing or how any of it was remotely applicable, Hurray for this guy!
I'm 70 years old and I have a daughter who is 5, almost 6. No problem so far but watching a few of your problems I find there's lots I don't remember from not using it since 76 I always enjoyed math in school. It's like doing puzzles. Wish I had had you as a teacher when I was young. Anyway this is going to be fun to relearn what I've forgotten and help my daughter as she grows.
I was proud that I got the right answer, the long way, though ... multiplying and employing the quadratic formula. I felt a little foolish when you made it so simple. Thanks for the videos.
Not sure if this is just dumb luck, but here is how I got it. This is basically vertex form, we just need to clean it up by factoring out the 3 on the left and setting the whole thing equal to zero. which makes the equation 3(c - 2/3)^2 - 14. This is the form of a(x-h)^2 + k x = h + sqroot of (-k) / a and x = h - sqroot of (-k) / a plug in values you get c = 2/3 + sqroot of (14) / 3 for the first solution and 2/3 - sqroot of (14) / 3 for the second solution. Note that we are only taking the square root of the top not the bottom in both formulas. a little complicated to understand since this is written on computer, but if you can visualize it on a board it makes sense.
I would suggest the following solution: (3c-2)^2=14 | -14 ad rewrite 14 as sqrt(14)^2 (3c-2)^2--sqrt(14)^2=0 | 3. binomic fomula ((3c-2)-sqrt(14)*((3c2)+sqrt(14))=0 A product has the value 0,if and onl if att least 1factor has the value 0: 3c-2-sqrtt(14)=0 or 3c-2+sqrt(14)=0 3c=2+sqrt(14) or 3c=2-sqrt(14) c=(2+sqrt(14))/3 or c=(2-sqrt(14))/3 In my opinion, this is the better wa to solve this equation. ou dont need to work witth +-sqrt(), you simpl hae a productt with value 0,if ou do ittt this way. For solution, you have only to sett each factor to 0. You get the same soluttions,buttt witthouttt tthe "+-sqrtt()" thing.
Don't forget that there are 2 solutions for c (quadratic equation) (I mean: take the square root of both sides and you get 3c - 2 = +/- sqr. rt. 14 and the rest is easy).
I was on the right track but added an unnecessary step at the end dividing 2 + the square root of 14 by 3 which gave me 2 + the square root of 14 over 9. :( And I forgot that there was 2 solutions also. But I'm happy that I saw what to do at the beginning. Moral victory! :)
Quadratic equation, that's my thing: First change c into x because of the c in the ABC-formula. 9x² - 12x + 4 = 14 so the equation will be y = f(x) = 9x² - 12x - 10 = 0 with a = 9 b = -12 and c = -10 Discriminant D = b² -4ac = (-12)² - 4 . 9 . (-10) = 144 + 360 = 504 ( positive so 2 solutions for y=0 ) Delta d = SQR ( D ) / 2a = 6V14 / 2.9 = V14/3 (distance between xtop and y=0) X-coordinate of the parabola xtop = - b / 2a = - (-12) / 2.9 = 12/18 = 2/3 So x = xtop minus/plus delta = 2/3 -/+ (V14)/3 = ( 2 -/+ V14) / 3 And to finish this up: ytop = f(xtop) = 9 . ( 2/3 )² - 12 . 2/3 - 10 = 4 - 8 -10 = - 14 So the three points of the parabola are: x1( (2-V14)/3 , 0 ) xtop ( 2/3 , -14 ) and x2 ( (2+V14)/3 , 0 )
Your videos are intriguing. It has been ages since did this. I remember somethings and not others. The refresher a moments where it all falls in place. Great fun.
Yes, the equation is already in the (h,k) format and can be manipulated directly without any further work. But I do believe its a good exercise for new learners to see the different methods that can be used to solve quads, when I was learning for the first time, I really struggled with quadratic equations, now I see many different routes to the solution, I believe this was made possible by using all the different methods and repetition, as my fluid Inelligence is lacking. Therefore, even though the fastest and most efficient route is simple linear manipulation, going through the whole gambit of putting the equation back into ax^2+bx+c format and completing the square is good practice for those of us less gifted, like myself. I'm sure this problem is chilsplay to you and many others, but to new learners, it's a perfect jumping off point, I used the omplete the square method, as I loathe the quadratic formula, not sure why. Thanks for the content fella
I've never had much luck with the quadratic equation. I wrote it down on a cue card for a math exam, but when I tried to use it, I couldn't find it, but got in trouble for fidgeting during the test! I said I had a bad itch. Then later I found it in the washing machine, all shredded!
Math Skills rebuilder I graduated in 76 No Calculators We have tables in the back of the book Dose the rebuilder course have tables???? Yeh I used a slide rule My senior year The rich kids had SR-31 calculators I believe they cost 200 dollars When I went to college the price came down to 30 dollars That happened with in a year Calculators today can do your graphing for crying out loud I had to use LOG tables and Trig tables in High School Algebra 2 and College Algebra Not in College Trig or Calculus is was all calculators I can’t remember when the graphing calculators came out I didn’t use one in trig or Calculus Lol I took computer programming and our Language was Fortran 4 Yeh I am showing my age The computer was a main Frame that we called into by telephone we had a keyboard and a printer no cathode Ray Tune ( Monitor) which is like a small TV No computer screen one keyboard one printer for 20 students Ohh what fun it was That was cutting edge back then .
I took Fortran IV in Jr college, and the mainframe geek was all upset with their new high speed printer, when wise guys stuck a card into their stack punched with 1H1 ( one Hollerith one ) on it, because that made the printer spew out paper like Lucille's Chocolate factory. Then he had to clean up the piles of ruined paper stock, then he ran out of paper, but the admin wouldn't let him place orders for more paper as that was over budget. Those were the days. We'd laugh about it at lunch.
When I was learning this stuff, everyone said, "nobody is going to use this in real life". Now that I'm finished with my working life and enjoying retirement, I can safely say that that was 100% correct! My career was environmental science.
(3c - 2)^2 = 14 The important thing to remember here is that (3c - 2)^2 is NOT just 3c^2 + 4 and (3c - 2)(3c - 2) = 9c^2 - 6c - 6c + 4 doesn't really ''make this EASY!'' So let's just do the square root of both sides .. 3c - 2 = the square root of 14 3c = the square root of 14 + 2 c = (the square root of 14 + 2) / 3 and I've hit another dead end here 🤔🤔!!
John, every time you explain taking the square root of a number I get confused about the + and - part. You say if it's the "principle" square root, then only use + sign in front of the answer, not the - sign. However, in this case, your answer leaves the "14" under the sq. rt. sign (i.e. you did NOT find the sq. rt. of 14, but left 14 under the radical sign. Yet you put a +/- in front of the radical sign. Why? . . . Let's say the sq. rt of 14 is 3.8. THEN you would use = (2) +/- 3.8. But you did not find the square root . . . you left the radical with 14 under it. Does the fact that this is a quadratic equation, then require the +/- sign? Whereas, in your example, if your problem is only finding the sq. rt. of 4, your answer is +2, not +/-2 .
An equation like x^2=c hasthe 2 solutions +sqrt(c) and -sqrt(c) (if c is bigger than 0). So you must keep care of both possibilities ... I would sugesst to use another way to solve this equation, where you have nott tto keep care of "+-sqrttt(...)": (3c-2)^2=14 | -14 ad rewrite 14 as sqrt(14)^2 (3c-2)^2--sqrt(14)^2=0 | 3. binomic fomula ((3c-2)-sqrt(14)*((3c2)+sqrt(14))=0 A product has the value 0, if and onl if att least 1 factor has the value 0: 3c-2-sqrtt(14)=0 or 3c-2+sqrt(14)=0 3c=2+sqrt(14) or 3c=2-sqrt(14) c=(2+sqrt(14))/3 or c=(2-sqrt(14))/3
@@juergenilse3259 Our assertion is that "c=(2-sqrt(14))/3" is mathematically false because it allows 1 = -1. The Negative can be a number, but CANNOT be an operation.
@@Stone_624 Whh do ou think, it allows 1=-1 ? It is just 1 of the 2 solutions in the video (even if iit is noticed a little bit different). I have wrote both solutions separat(with "or" between them)instead of the notationused in the video. Nothing in m solution allows 1=-1. And i see noreason, wh it should be mathemahecall incorrect.
@@juergenilse3259 You don't understand the two different "Equivalences" I'm talking about. They're the same NUMBERS, but they're WRITTEN differently. The issue is that the way they're written is NOT exactly the same as the previous step, even though "Mathematically" they both equal the same numbers. This is a very high level of mathematics (3-4 levels past Calculus is when you start getting into these types of mathematical nuances) In programming, there's different types of Data. You can have an Integer 1, and a Float 1.0, as well as a String "1" . Now Clearly 1 == 1 == 1 . One is Equal to 1, whether the 1 is an Integer data type, or a Float data type, or represented as a string. This equivalence is usually represented by two equal signs in programing 1 == 1.0 is true. 1 == "1" is true. In this example, You're saying 2 - SQRT(14) == 2 + (-SQRT(14)) . MATHEMATICALLY, This is correct. But I'm saying SEMANTICALLY it's not correct. Back to the programming, we have a "Stricter" form of Equivalence, usually written with Three equal signs. This equivalence asserts that the Data Type also matches, not just the data itself. If you do a Stricter equivalence, Integer 1 === Float 1.0 is FALSE . integer 1 === String "1" is FALSE, Because the underlying representation doesn't match, even though what they're representing is equivalent. What We're saying is that under a "Stricter" form of equivalence, The Minus operation is incorrect, Because you're saying "2 + SQRT(14) == 2 - SQRT(14)" =(-2 both sides)=> "+ SQRT(14) == - SQRT(14)" =(divide by SQRT(14) both sides)=> "1 == - 1" . Now Mathematically this IS kinda-sorta correct because the SQRT(14) EVALUATES to two answers, which cancel out the operations for both SQRT, sort of collapsing down into a duel-equivalence (and also duel-wrong-ness). But without explicitly accounting for that in a semantic way (By using additional parenthesis to express the +- is part of the SQRT(14) Evaluation, NOT the operation, and explicitly writing the + to inform a single objective operation, Not two separate operations, one of which is semantically incorrect but mathematically correct), You're opening yourself up to a mathematical fallacy and a mathematically valid argument that the answer is incorrect, despite (counterintuitively) it being mathematically equivalent. This is actually why I HATE subtraction as an operation. Subtraction doesn't need to exist. All subtraction can be reduced to Addition of a Negative, Which is far more mathematically sound in every situation.
The +/- means 2 values for c: For + : (2+√14) ÷3 it's 1.91389... For - : (2-√14) ÷3 it's -3.741657386... They both work, because of (3c-2)^2 Check ot out!
I clicked on this video because I wanted to know if I could do anything with 14^(1/2). A bit disappointed, But couldn't find anything myself searching before watching. Also from a purely syntactical point of view I would personally disagree with your "+- SQRT(14)" assumption there. The RESULT of SQRT(14) is + and - , But putting the operation +- BEFORE (outside) of the SQRT only works because the result of the SQRT has 2 answers, + and - , ie because of a technicality, or by coincidence. If you would EVALUATE the SQRT to give a + and - answer, that'd be OK, But you're saying basically that +SQRT(14) === +-SQRT(14) , which TECHNICALLY (Syntactically) isn't true, Because that implies +1= -1, which is objectively false. You need to wrap the +-SQRT(14) in Parenthesis to imply that it's a calculation of the original SQRT(14), Implying the Outside of that expression is explicitly positive, which resolves the mathematical ambiguity, by moving the +- from separate operations (2 operations on 1 number) into being part of the number that the operation is acting on (1 operation on 2 numbers). In other words I'm saying that the "2 - SQRT(14)" part is expressly false, and mathematically not equivalent to the original equation. I would assert the correct answer is (2 + (+-SQRT(14))) / 3 , As opposed to the (2 +- SQRT(14)) / 3 you describe here. This explicitly resolves the 1 = -1 issue.
Well, if this one doesn't grab you, go for Rod Serling's impromptu dictation technique. If we didn't have film proof of his existence nobody would believe it.
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This guy is the bomb!! He has such an easy, affable delivery that I think I could have learned to love math had he been my mentor 'back in the day'. I understood the mechanics back then, but never the 'why?' of what I was doing or how any of it was remotely applicable, Hurray for this guy!
I'm 70 years old and I have a daughter who is 5, almost 6. No problem so far but watching a few of your problems I find there's lots I don't remember from not using it since 76 I always enjoyed math in school. It's like doing puzzles. Wish I had had you as a teacher when I was young. Anyway this is going to be fun to relearn what I've forgotten and help my daughter as she grows.
I was proud that I got the right answer, the long way, though ... multiplying and employing the quadratic formula. I felt a little foolish when you made it so simple. Thanks for the videos.
Not sure if this is just dumb luck, but here is how I got it.
This is basically vertex form, we just need to clean it up by factoring out the 3 on the left and setting the whole thing equal to zero. which makes the equation 3(c - 2/3)^2 - 14. This is the form of a(x-h)^2 + k
x = h + sqroot of (-k) / a and x = h - sqroot of (-k) / a
plug in values you get
c = 2/3 + sqroot of (14) / 3 for the first solution and 2/3 - sqroot of (14) / 3 for the second solution. Note that we are only taking the square root of the top not the bottom in both formulas.
a little complicated to understand since this is written on computer, but if you can visualize it on a board it makes sense.
You seem to know this but it's not making sense to me: what is k? Does k have some value? What is vertex? Does vertex have an application?
I see, k is -14.
What is vertex used for?
I can't edit my post. I see your values for c are identical to mine! Thanks.
Wait!!!! How can you take the square root of a negative number (-k)?
Is this "imaginary numbers?"
I would suggest the following solution:
(3c-2)^2=14 | -14 ad rewrite 14 as sqrt(14)^2
(3c-2)^2--sqrt(14)^2=0 | 3. binomic fomula
((3c-2)-sqrt(14)*((3c2)+sqrt(14))=0
A product has the value 0,if and onl if att least 1factor has the value 0:
3c-2-sqrtt(14)=0 or 3c-2+sqrt(14)=0
3c=2+sqrt(14) or 3c=2-sqrt(14)
c=(2+sqrt(14))/3 or c=(2-sqrt(14))/3
In my opinion, this is the better wa to solve this equation. ou dont need to work witth +-sqrt(), you simpl hae a productt with value 0,if ou do ittt this way. For solution, you have only to sett each factor to 0. You get the same soluttions,buttt witthouttt tthe "+-sqrtt()" thing.
Nothing cranks the brain like a little math exercise in the morning with my coffee.👍
WOW. That's dedication!
Don't forget that there are 2 solutions for c (quadratic equation) (I mean: take the square root of both sides and you get 3c - 2 = +/- sqr. rt. 14 and the rest is easy).
Thank you for the training! You make it seem easy! 🙏🙏
I was on the right track but added an unnecessary step at the end dividing 2 + the square root of 14 by 3 which gave me 2 + the square root of 14 over 9. :(
And I forgot that there was 2 solutions also.
But I'm happy that I saw what to do at the beginning. Moral victory! :)
Quadratic equation, that's my thing:
First change c into x because of the c in the ABC-formula.
9x² - 12x + 4 = 14 so the equation will be y = f(x) = 9x² - 12x - 10 = 0 with a = 9 b = -12 and c = -10
Discriminant D = b² -4ac = (-12)² - 4 . 9 . (-10) = 144 + 360 = 504 ( positive so 2 solutions for y=0 )
Delta d = SQR ( D ) / 2a = 6V14 / 2.9 = V14/3 (distance between xtop and y=0)
X-coordinate of the parabola xtop = - b / 2a = - (-12) / 2.9 = 12/18 = 2/3
So x = xtop minus/plus delta = 2/3 -/+ (V14)/3 = ( 2 -/+ V14) / 3
And to finish this up: ytop = f(xtop) = 9 . ( 2/3 )² - 12 . 2/3 - 10 = 4 - 8 -10 = - 14
So the three points of the parabola are: x1( (2-V14)/3 , 0 ) xtop ( 2/3 , -14 ) and x2 ( (2+V14)/3 , 0 )
Seriously?!?!
Super love math problems the way other people love crossword puzzles. Thank you for this fantastic channel!
Your videos are intriguing. It has been ages since did this. I remember somethings and not others. The refresher a moments where it all falls in place. Great fun.
Thanks. I needed that!
Yes, the equation is already in the (h,k) format and can be manipulated directly without any further work. But I do believe its a good exercise for new learners to see the different methods that can be used to solve quads, when I was learning for the first time, I really struggled with quadratic equations, now I see many different routes to the solution, I believe this was made possible by using all the different methods and repetition, as my fluid Inelligence is lacking. Therefore, even though the fastest and most efficient route is simple linear manipulation, going through the whole gambit of putting the equation back into ax^2+bx+c format and completing the square is good practice for those of us less gifted, like myself. I'm sure this problem is chilsplay to you and many others, but to new learners, it's a perfect jumping off point, I used the omplete the square method, as I loathe the quadratic formula, not sure why. Thanks for the content fella
I've never had much luck with the quadratic equation. I wrote it down on a cue card for a math exam, but when I tried to use it, I couldn't find it, but got in trouble for fidgeting during the test! I said I had a bad itch. Then later I found it in the washing machine, all shredded!
(3c -2) sq = 14 => 3c-2 = sqrt 14 => 3c = 2 + sqrt 14 => c = (2 + sqrt 14) / 3 and because "c" was an unknown, plus and minus sqrt 14
That was a great quadratic equation exercise.
You Explain very well! Please bring your audio up. Thanks!
Turn your volume up...thanks
@@facepvh
. WHAT?
It's more about being able to recognize which solution works best.
I always enjoyed algebra in school but I must admit I really never used it in my work but I did use Trigonometry and geometry.
Math Skills rebuilder I graduated in 76 No Calculators We have tables in the back of the book Dose the rebuilder course have tables???? Yeh I used a slide rule My senior year The rich kids had SR-31 calculators I believe they cost 200 dollars When I went to college the price came down to 30 dollars That happened with in a year Calculators today can do your graphing for crying out loud I had to use LOG tables and Trig tables in High School Algebra 2 and College Algebra Not in College Trig or Calculus is was all calculators I can’t remember when the graphing calculators came out I didn’t use one in trig or Calculus Lol I took computer programming and our Language was Fortran 4 Yeh I am showing my age The computer was a main Frame that we called into by telephone we had a keyboard and a printer no cathode Ray Tune ( Monitor) which is like a small TV No computer screen one keyboard one printer for 20 students Ohh what fun it was That was cutting edge back then .
Mr John thanks for your history
Did you keypunch cards to give the mainframe geek?
I took Fortran IV in Jr college, and the mainframe geek was all upset with their new high speed printer, when wise guys stuck a card into their stack punched with 1H1 ( one Hollerith one ) on it, because that made the printer spew out paper like Lucille's Chocolate factory. Then he had to clean up the piles of ruined paper stock, then he ran out of paper, but the admin wouldn't let him place orders for more paper as that was over budget. Those were the days. We'd laugh about it at lunch.
Did you ever use a Keuffel & Esser slide rule? Those babies are works of ART!
got it. took sq rt and solved. thanks for the fun
When I was learning this stuff, everyone said, "nobody is going to use this in real life". Now that I'm finished with my working life and enjoying retirement, I can safely say that that was 100% correct! My career was environmental science.
Conclusion: environmental science isn’t science, its a religion.
Never used it as a research chemist. Statistcs, yes. Algebra, no.
Never used it in carpentry, but I sure tried to. Now I use it to get to sleep!
Good stuff A Quadratic Equation
I’m bad at math. I thought it was c was 3
But the way I did it was 14 divided by 2 is 7
So 7 plus 2 is 9 and 3 goes into nine 3 times.
SO c is 3
(3c - 2)^2 = 14
The important thing to remember here is that (3c - 2)^2 is NOT just 3c^2 + 4
and (3c - 2)(3c - 2) = 9c^2 - 6c - 6c + 4 doesn't really ''make this EASY!''
So let's just do the square root of both sides ..
3c - 2 = the square root of 14
3c = the square root of 14 + 2
c = (the square root of 14 + 2) / 3 and I've hit another dead end here 🤔🤔!!
Of course if you look at the pattern of the answer you notice the quadratic equation.
How are QE used in life. I sincerely want to understand why we have them.
C=(-+√14. +2)/3
Simple algebra will solve this equation. No quadratic formula is needed. Just keep in mind that SQRT (14) can be + or -
John, every time you explain taking the square root of a number I get confused about the + and - part. You say if it's the "principle" square root, then only use + sign in front of the answer, not the - sign. However, in this case, your answer leaves the "14" under the sq. rt. sign (i.e. you did NOT find the sq. rt. of 14, but left 14 under the radical sign. Yet you put a +/- in front of the radical sign. Why? . . . Let's say the sq. rt of 14 is 3.8. THEN you would use = (2) +/- 3.8. But you did not find the square root . . . you left the radical with 14 under it.
Does the fact that this is a quadratic equation, then require the +/- sign? Whereas, in your example, if your problem is only finding the sq. rt. of 4, your answer is +2, not +/-2 .
An equation like x^2=c hasthe 2 solutions +sqrt(c) and -sqrt(c) (if c is bigger than 0). So you must keep care of both possibilities ... I would sugesst to use another way to solve this equation, where you have nott tto keep care of "+-sqrttt(...)":
(3c-2)^2=14 | -14 ad rewrite 14 as sqrt(14)^2
(3c-2)^2--sqrt(14)^2=0 | 3. binomic fomula
((3c-2)-sqrt(14)*((3c2)+sqrt(14))=0
A product has the value 0, if and onl if att least 1 factor has the value 0:
3c-2-sqrtt(14)=0 or 3c-2+sqrt(14)=0
3c=2+sqrt(14) or 3c=2-sqrt(14)
c=(2+sqrt(14))/3 or c=(2-sqrt(14))/3
Did you actually beat me to this? Finally, a Worthy Opponent! (See my comment)
@@juergenilse3259 Our assertion is that "c=(2-sqrt(14))/3" is mathematically false because it allows 1 = -1. The Negative can be a number, but CANNOT be an operation.
@@Stone_624 Whh do ou think, it allows 1=-1 ?
It is just 1 of the 2 solutions in the video (even if iit is noticed a little bit different). I have wrote both solutions separat(with "or" between them)instead of the notationused in the video. Nothing in m solution allows 1=-1. And i see noreason, wh it should be mathemahecall incorrect.
@@juergenilse3259 You don't understand the two different "Equivalences" I'm talking about. They're the same NUMBERS, but they're WRITTEN differently. The issue is that the way they're written is NOT exactly the same as the previous step, even though "Mathematically" they both equal the same numbers. This is a very high level of mathematics (3-4 levels past Calculus is when you start getting into these types of mathematical nuances)
In programming, there's different types of Data. You can have an Integer 1, and a Float 1.0, as well as a String "1" . Now Clearly 1 == 1 == 1 . One is Equal to 1, whether the 1 is an Integer data type, or a Float data type, or represented as a string. This equivalence is usually represented by two equal signs in programing 1 == 1.0 is true. 1 == "1" is true. In this example, You're saying 2 - SQRT(14) == 2 + (-SQRT(14)) . MATHEMATICALLY, This is correct. But I'm saying SEMANTICALLY it's not correct.
Back to the programming, we have a "Stricter" form of Equivalence, usually written with Three equal signs. This equivalence asserts that the Data Type also matches, not just the data itself. If you do a Stricter equivalence, Integer 1 === Float 1.0 is FALSE . integer 1 === String "1" is FALSE, Because the underlying representation doesn't match, even though what they're representing is equivalent.
What We're saying is that under a "Stricter" form of equivalence, The Minus operation is incorrect, Because you're saying "2 + SQRT(14) == 2 - SQRT(14)" =(-2 both sides)=> "+ SQRT(14) == - SQRT(14)" =(divide by SQRT(14) both sides)=> "1 == - 1" . Now Mathematically this IS kinda-sorta correct because the SQRT(14) EVALUATES to two answers, which cancel out the operations for both SQRT, sort of collapsing down into a duel-equivalence (and also duel-wrong-ness). But without explicitly accounting for that in a semantic way (By using additional parenthesis to express the +- is part of the SQRT(14) Evaluation, NOT the operation, and explicitly writing the + to inform a single objective operation, Not two separate operations, one of which is semantically incorrect but mathematically correct), You're opening yourself up to a mathematical fallacy and a mathematically valid argument that the answer is incorrect, despite (counterintuitively) it being mathematically equivalent.
This is actually why I HATE subtraction as an operation. Subtraction doesn't need to exist. All subtraction can be reduced to Addition of a Negative, Which is far more mathematically sound in every situation.
The time I spent learning algebra in high school would have been better spent learning entrepreneurship and personal finance.
i always thought whata in brackets cane first, in this case 3c-2 =1c
It's 3c - 2 NOT 3c - 2c
@@MrMousley been 60 years since i left school so l have forgotten most of the rules. i still dont understand algebra very well
How can 3c-2 = 1c?
@@debeeriz
It's never too late!
John can lift your spirits!
SEMPER FI!!
@@btpcmsag 3Cans of beer take 2 away and you have 1 can of beer
I got 1.91
The +/- means 2 values for c:
For + : (2+√14) ÷3 it's 1.91389...
For - : (2-√14) ÷3 it's -3.741657386...
They both work, because of (3c-2)^2
Check ot out!
Now goet a calculator to solve your answer.
This is wrong the answer is 1.91
This gives two values for f(c) = 14
For + √14 it's c = 1.91389...
For - √14 it's c = -3.741657386...
Try them in (3c-2)^2 ... They work!!
Roots - plus 2 - divided by 3. That's it. A three step solution to explain in 30 seconds. Boring...
I clicked on this video because I wanted to know if I could do anything with 14^(1/2). A bit disappointed, But couldn't find anything myself searching before watching.
Also from a purely syntactical point of view I would personally disagree with your "+- SQRT(14)" assumption there. The RESULT of SQRT(14) is + and - , But putting the operation +- BEFORE (outside) of the SQRT only works because the result of the SQRT has 2 answers, + and - , ie because of a technicality, or by coincidence. If you would EVALUATE the SQRT to give a + and - answer, that'd be OK, But you're saying basically that +SQRT(14) === +-SQRT(14) , which TECHNICALLY (Syntactically) isn't true, Because that implies +1= -1, which is objectively false.
You need to wrap the +-SQRT(14) in Parenthesis to imply that it's a calculation of the original SQRT(14), Implying the Outside of that expression is explicitly positive, which resolves the mathematical ambiguity, by moving the +- from separate operations (2 operations on 1 number) into being part of the number that the operation is acting on (1 operation on 2 numbers).
In other words I'm saying that the "2 - SQRT(14)" part is expressly false, and mathematically not equivalent to the original equation.
I would assert the correct answer is (2 + (+-SQRT(14))) / 3 , As opposed to the (2 +- SQRT(14)) / 3 you describe here. This explicitly resolves the 1 = -1 issue.
See my earlier post/question to John - I also questioned why he has +/- sqrt 14 instead of just +sqrt 14.
The world isn't going to start using two more sets of parentheses just on account of your tantrum. . . Sorry.
@@terry_willis
There are two solutions and both work.
I hate math
Well, there's always the Dirty Jokes videos.
Well, if this one doesn't grab you, go for Rod Serling's impromptu dictation technique. If we didn't have film proof of his existence nobody would believe it.