10% of Math Students can solve this equation
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- Опубликовано: 16 сен 2024
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By observation, m=-1 is an answer. Since the function is strictly increasing, it must be the only answer. Done.
As square roots are always positive and > 0 only 1 is left as logical conclusion
@@tigerdancer61 not really unless you know that they are not rational unless integers which lets only 1+1=2…. Well…. I guess…. Tell me
@@tigerdancer61 What is the square root of 0?
I used the same idea as u did, nice
I think it's easier to square both sides of the equation, then bring the part which has a square root to one side of the equation then square again to get rid of the square root and then complete the square.
2=√(5m+6)+√(3m+4)
2^2=(√(5m+6)+√(3m+4))^2
4=(5m+6)+2√(5m+6)√(3m+4)+(3m+4)
4=2√((5m+6)(3m+4))+8m+10
4-8m-10=2√((5m+6)(3m+4))
-8m-6=2√((5m+6)(3m+4))
-4m-3=√((5m+6)(3m+4))
(-1)(4m+3)=√((5m+6)(3m+4))
(-1)^2*(4m+3)^2=(5m+6)(3m+4)
(4m+3)^2=15m^2+(4*5+6*3)m+24
16m^2+24m+9=15m^2+38m+24
(16-15)m^2+(24-38)m=24-9
m^2-14m=15
m^2-14m-15=0
(complete the square or use the quadratic formula)
That is how I did this
How'd you get the sqrt symbol in a comment ?
@@agnichatian option+V
I thought it made sense also to square both sides initially, seeing this would produce a single square root term. Then, isolate the square root to one side and square again anticipating some sort of quadratic. I'd hate to rely on guessing or on integer answers. But these exercises often are designed around them.
Upon inspection the lhs is clearly strictly increasing and exists over the reals only for x>-6/5 where the lhs takes a value sqrt(0.4)
Graphically, the only solution there is -1. Rearranging to 34m^2 + 84m + 50 = 0 and then fed into the Brahmagupta quadratic formula, gives the two solutions, -25/17 and -1. This is in agreement with your statement.
We shouldn't include higher mathematics in this 9th grade priblem no doubt you may be much more skilful than a guy who made vdo but we should try to keep things simple rather than distracting audiences by higher knowledge
m=-1, by inspection. Took 5 seconds. (Each root has to be positive - all the hint you need.)
Good observation…
The point of it is to see who will get caught up with the standard rules of mathematics which doesn't always work. LOL. I found that out when I took physics. You must also consider the condition. 5m+6 = 1. Solve for m you get -1. Right. You plug that into the condition, and it satisfies it. LOL 1+1 = 2. The trick is in noticing that the square root of 1 is still 1. ha ha. It's funny. Admit it!
I think it is easier to just square both side of the initial equation, collect like terms and simplify, put the single radical expression on one side, then square both sides again, and you get a quadratic to solve.
You have a math brain. PLEASE AVOID PHYSICS. LMAO
No brother the solution get longer as still v have a quadratic with radical to solve by squaring the initial eqn as the given solution ft in all type of student so its makes more versatile as may be you n me finds it easier but when we are in teaching we have to think about the learning skills of all types of students so thats the point❤
@@MrBhatt1971 The solution is m = -1
@@MrBhatt1971 The trick to understanding physics is in realizing that the standard methods of mathematics doesn't always apply. Sometimes you must evaluate the condition, and then adjust the mathematics to fit it.
@@MrBhatt1971 Do me a favor. Forget about the rules. Here's how I approached the problem: I said, "for what value of m can I get a number such that the sum will equal 2. You take the binomials inside the radicals and you set one of them equal to 1. Then, you'll discover the answer. This is a trick question.
It didn’t help that my Algebra 1 teacher wasn’t a very nice person. That always makes it harder to learn. Kudos for being such a kind teacher!
This is the kind of question in workbook, if exam that is another level always lol!
You can get 15 to work in the original equation. Arbitrarily take the negative value of Sqrt(49). So you get Sqrt(81)=9, sqrt(49)=-7, 9 + (-7) =2.
Me too.
Once I got 5m+6 = 4 - 4√(3m+4) + (3m+4),
I used the substitution u=√(3m+4)
In this way the first member can be rewritten as follow
5m+6 = (5/3)(3m + 4 - 2/5) = (5/3)(u²-2/5).
And the equation become:
(5/3)(u²-2/5)=4 - 4u +u²
(5/3-1)u² -4u - 2/3 - 4=0
(2/3)u² - 4u - 14/3=0
Multiplying both sides by 2/3 we'll get
u² + 6u - 7 = 0
(u+7)(u-1) = 0
Which has two valid solutions: u=1 and u=-7.
So √(3m+4) = 1 lead us to m=-1, and that's OK.
But it is also true that √(3m+4)=-7 whose solution m=15.
I can't wy we should throuw away ths solution.
Isen't it true that √(3•15+4) = √(45+4) = √49 = -7?
If so, m=15 is valid too.
"Arbitrarily take the negative value of Sqrt(49)."
The square root of a non-negative real number is _defined_ to be the non-negative number whose square gives the original number. So taking the negative value here would contradict the definition of "square root".
@@bjornfeuerbacher5514 Agreed!
Interesting !
I succeeded by an equivalent method. At the beginning I square the 2 sides. Calculate then square a second time, which gives the same trinomial
Before watching the video:
Given: √(5m + 6) + √(3m + 4) = 2
To find: m
Moving the second square root term to the RHS:
√(5m + 6) = 2 - √(3m + 4)
Squaring both sides:
5m + 6 = 4 + 3m + 4 - 4√(3m + 4)
Moving all non-square-root terms to the LHS and simplifying:
2m - 2 = -4√(3m + 4)
Dividing through both sides by 2:
m - 1 = -2√(3m + 4)
Squaring both sides again:
m² - 2m + 1 = 12m + 16
Moving all terms to LHS and simplifying:
m² - 14m - 15 = 0
As the sum of coefficients is 0, m = - 1 is a solution.
Using this fact to factorise the quadratic:
(m + 1)(m - 15) = 0
From this, m can be -1 or 15. However, m = 15 is clearly an extraneous solution from the second squaring step, and can be ignored.
This can be confirmed by plugging into the first suqare root term in the original equation which gives us 9 + a positive term (square root is always positive), which can never be 2.
Thus, m = -1.
Can you explain this bit because that's what I still don't get... "Using this fact to factorise the quadratic: (m + 1)(m - 15) = 0"
What is factorising a quadratic?
@@phugoid if the sum of the coefficients of ax² + bx + c, ie, a + b + c = 0, then x = -1 is always a root of the polynomial. This works for any degree n, not just 2. It has a formal name, but it's escaping me at the moment.
@@phugoid any quadratic ax² + bx + c can be written as (x - m)(x - n) where m and n are complex numbers, and the roots to the quadratic, i.e., x = m and x = n are the solutions to the original quadratic.
m and n being roots of the quadratic means that am² + bm + c = an² + bn + c = 0, btw.
This is generally true of any polynomial of any integer degree
(The degree of the polynomial is the highest power of the variable term.)
I only ever took high-school level math myself, so I'm not very good at explaining this.
(x - m) and (x - n) are called factors of the polynomial, as they multiply together to give the original polynomial back, much as factors of a compound number multiply together to give the original number back. So factorising a polynomial is separating it into a series of lower degree terms being multiplied together. It's useful when finding solutions to a quadratic, as the two linear factors can be independantly equated to zero to solve for the variable term.
When I looked at the equation and saw the answer “2” I knew the radicles had to equal 1 such that 1 + 1 = 2. So then I thought “what number when multiplied by 5 and then added to 6 would give me the result of 1.” The only way I could get a result of 1 was to make 5 a negative 5 and the only way to make 5 a negative 5 was to multiply it by negative 1 (-1). Then I asked myself if -1 would also work in the second radicle. It does so I knew the answer was negative 1 (-1). Of course this was a very special equation and my method does not work in all cases, but it did here.
Indeed the answer is relatively obvious. I think it took me around 20 seconds. It was obvious m < 0. So I tried -1. It worked.
I noticed as well, but solved it formally also because of the possibility for additional solutions which weren't so obvious. It's dangerous to be satisfied to get an answer when it's possible there is more than one answer.
It could have been 1/2+3/2, or (3-i)+(-1+i), or any other myriad possibilities. 1+1 made sense, and worked, but it didn't necessarily have to be the case.
I think that,at the time 7'17 we should add the condition m-1
Suppose let a=sqrt(5m+6) , b=sqrt(3m+4)
Multiply both sides by a-b
gives a²-b² ==2(a-b)
rearranging: a²-2a=b²-2b
So a=b
ie sqrt( 5m+6)= sqrt(3m+4)
5m+6= 3m+4
2m=-2
m=1
Is there a flaw in the above?
sorry, m= - 1
@@siewkailo8788 you can edit you reply
Much better, in fact beautiful solution, modulo minus sign :-)
Yes. You are multiplying both sides by zero.
yes. if your proof were true then any a,b. such tjat a+b=2 would also find a=b
You can also multiply both sides by sqrt(5m+6)-sqrt(3m+4) and rearrange terms. You will get quadratic equation : m^2-14m-15=0
Which will give you m=-1 and 15.
Most of the people asserted that m=15 is not likely a solution, however if you assume the sign of sqrt can be +ve or -ve, that means you can accommodate this solution as well.
Hi Kedan,
It's best to test your solutions in the original equation. Doing so, you will find that m = 15 does not satisfy the original equation. Cheers!
When m = 15 LHS becomes 9 + 7 clearly not 2.
Also, when the format √y is used, the classical reading says that √y is always positive and never negative. Thus, even though 49 can be 7² or (-7)², √49 is always the principal value, which is 7. To also include -7, the proper way is to use fractional exponents. 49↑½ has to be both ±7.
@@stevenzimmerman3945 what decides when you can only take a positive square root?
@@GirishManjunathMusic classical reading? Is it really standard? In the wild \sqrt {49} can take on two values. What's so different about \sqrt{3m+4}? What field are we assuming we're in? If it's C we're guaranteed to have two roots
Guys, a Root of 49 can be 7 and -7, so 15 is also the correct answer,
sqrt (5m+6) + sqrt(3m+4) = sqrt(75 + 6) + sqrt (45 +4) = sqrt(81) +sqrt(49) = 9 +(-7) = 2
LHS =RHS @TableClass Math
no, roots are by definition positive numbers. Otherwise a number would have two roots, and sqrt would not be a function.
I did this by using number theory. If the sum of two positive square roots is equal to a positive integer each square under the radical must be a perfect square number. So sqrt of 5m+6 must equal 1 and sqrt of 3m + 4 must equal 1 because those are the only perfect squares that add up to 2 after taking the square root. Sometimes it pays to do the logic first and then the algebra. M=15 is a solution to a similar problem with the difference of two square roots. of perfect squares. So if that was the problem I'd know that m could only be an integer (no common factors between 3 and 5) and that each of the values under the radical's must be perfect squares. .
That would assume that m was an integer. What if it weren't?
I just squared everything.
One could have been 0 and the other sqrt 4.
@@MrLidless We could argue about 0 being a perfect square but if we assume 5m+6=0 and 3m+4=4 than their is no solution for m and the same holds true if we assume 5m+6=4 and 3m+4=0 .
@@kimobrien. Exactly. That should have been included in your answer to rule it out.
I was using mostly the same path of thinking, but in the end I used another formula.
The same until:
m^2-14m-15=0
m=(+14+-sqr( (-14)^2 - (4 * 1 * (-15))))/2
m=(14+-sqr(196+60))/2
m=(14 +- 16)/2
m=-2/2 or 30/2=> m=-1 or m=15...
Nice explanation of your work. Especially like the extraneous answers.
I have been teaching math for years, this is the first of your videos I clicked on. I would expect alot more students than 10% could easily see that you just square both sides, and then can solve it easily in your head.
And no, I know you want solve it in your head that way, just seems an easier approach. Nice video!
this method is only efficient if the solution is a non-integer that is hard to find randomly. in this case tho it's easy to make the deduction about LHS increasing and find the answer.
Ah, but surely 15 *is* a solution, you just accept that sqrt(3m-4) has *two* roots and accepting sqrt(49) can quite legitimately be -7, and yields the right answer...
Same reasoning here.
The clue of the whole thing is to bring the second square root to the right side of the equation. Then you get sqr.rt (5m + 6) = 2 - sqr.rt (3m + 4). The rest is easy.
I am a 72 year old retired electronics engineer. I have not used this type of mathematics since I left school in 1968. I could solve this by the method you outlined, but I would have used very a very different method to multiply the terms. What the heck is FOIL? What happened to straightforward long multiplication. No abstract rules to remember. Just stick one term underneath the other, multiply each element bottom to top, left to right and add them all together. Nothing to learn by rote. As for solving the quadratic? Not all quadratics factorise, so I would the use basic quadratic formula by default. I still remember that too. I must have had good maths teachers back in the day, I guess.
FOIL is acronym for First Outer Inner Last when one is multiplying a pair of binomials. I agree it is crutch that favors memorization over understanding. FOIL is distributive property applied twice.
Some US students would not know how to multiply a binomial by a trinomial because of the way algebra is taught in some places! Way to stay active in your retirement doing math!!
This is way overly too complex method.
This is an increasing function so
Just by checking n=1 you realize this is over 2, then n=0 over 2 as well, then n=-1 you find the solution, then n=-2 is impossible, so the only solution is n=-1
Good teacher. It's also handy that there are others out there who can do similar. I rate this video 10 out of 10
Finally, one of these that is so stupidly easy I can see the answer immediately.
If the total under each sq root is 1 and sq root(1) + sq root(1) = 2. So, 5m+6 =1 and 3m+4=1 OR 8m+10 = 2... m=-1
I m not a math freak and my abitur is almost 40 years ago, but I tried.
I squared it, seperared the roots to one side, squred again and got a simple quadratic equation with the solutions 15 and -1.
It was easy or I am genious?
I added the radicals and solved that way. I got the right answer -1. You solve it a different way
What do you mean by adding the radicals?
How do you add radicals? Please demonstrate!
I always liked to tease my students by giving them quadratic equations where x was replaced by some simple function, like x^3, Sqrt(1+x) or something like that. When asked, how to solve it, I would, reply, "what have we been studying?" They would say, "quadratic equations". Then I would answer, "maybe that is what it is..." They were mostly math-challenged business majors. And by now, all of them have climbed the corporate ladder and accumulated vast wealth, while I have not. :-)
The teaching of algrebra I/II, trig, geometry/analytical geometry, precalc and calc at the HS level needs a vast update -- the wort thing that even happened to HS math is AP courses which force a wide coverage but narrow depth solving to pass the test instead of a basic treatment of the subject on a limited range of subjects but which cover a large percentage of problems that you would actually see in a career.
@@haroldh3863 The word "career" is a fraud: it puts the student at the mercy of evil human resource managers. In fact, there is no "career"; only a future of exploitation followed by massive layoffs. Remember 1963 when Skybolt was cancelled and Douglas Aircraft laid off 3000 engineers in one day? Remember the Nixonian holocaust when Manned Orbital Lab, Fast Deployment Logistics Ship, 3 remaining Apollo missions, Supersonic Transport, and more were all cancelled at about the same time? Why not teach students a love of mathematics, logic, and problem solving that they can enjoy for life, no matter which Starbucks or McDonald's employs them part time? It sure beats beer, pot, and football. Yes, there are job openings... for those with a current SECRET security clearance (or family connections or blackmail information on upper management). Yes, there are good engineering jobs-- with the utility companies and the state highway departments. Math not required.
Also, there is a dot in the second term on the left side that makes it look like Sqrt(3*m + .4).
@@haroldh3863 I first had algebra in grade 9. My mother said, "What? I had it in grade 7! Are kids getting stupider?" She got as far as basic geometry, but never got to trig or calculus. In college at UC Berkeley she was a French and Spanish major, graduating in 1936.
@@richardnineteenfortyone7542 that's right in line with myself.. I had algebra 1 in 7th, geometry, then algebra 2, then trig, then precalc and then calc.. then I had two semesters engineering math, 3 semesters cal 1,2,3 diff Q, advanced engineering math, vector analysis and then engineering statistics. Add in some really math intensive classes like modern comms (Fourier, convolution etc), fields and waves ( all vector calc) and control systems (laplace, z transforms etc). I have had a long career in engineering which I am getting towards the end.. looking back gives me a perspective that few teachers have...I have done the academic and the practical, theory and application...I have also taught my children and others where I count several engineers and a physicist.
I dare say that, in my view, 15 too is a solution since -7 is also a square root of 49.
Iam 62... retired engr... but not tired refreshing math... mathematics is the best mental exercise... thanks...
I now have a massive head ach
We can also solve by taking 3m+4 as any constant so that we can solve easily
When you square root a number, the result is positive and negative. The m=15 does work if the first number is + and the second root is negative. It is a solution with limitations.
The square roots of 49 are the solutions to x^2 = 49 and therefore -7 and +7.
But the equation uses radixes, and the radix is by definition the principal (positive) square root of a real number. Using sqrt as calculator command is a poor choice of words causing the misunderstanding.
First observation is that the LHS is a strictly increasing function of m. Also, for m=0, the LHS is already too large, so try -1, which works. Are there other solutions? No. The LHS is strictly increasing, so it will equal 2 at exactly one point.
If -1 would not have worked, you could keep decreasing until the LHS was smaller than 2, and you would have bracketed the solution. Binary chop,or interpolation would keep improving the solution until you are arbitrarily close .
@
Jim Coyle Good thinking. Also, if we consider negative square roots then 9 - 7 = 2 which is a solution.
I think M could still equal 15. (square root of 81=9 ) + (square root of 49=-7)= 2.
That's not correct. The negative sign would have to be before the radical sign, such that: (sqrt 81) - (sqrt 49) = 9 - 7 = 2. But that would change the operation of the equation.
A negative number inside of the radical operator cannot yield a real number solution, but can only yield a complex number solution: sqrt(-49) = 7i ...
@@ministryoftruth8523 There was neither a positive nor a negative sign before either square root. Therefore either the negative or positive root could be used. Not sure where you got the idea of doing the square root of negative 49. I certainly wasn't talking about doing that. I was merely saying the that the square root of positive 49 could be either negative 7 or positive 7.
@@warblerab2955 I was just making the absolute point that when taking the square root of ANY number, the solution is NEVER a negative number. By the rules of mathematics: The square root of a negative number is not defined in the real number system. I don't know if you're currently studying math, but I assure you that if in a test you write your solution as: (sqrt a) = +/-b, you will, at best, receive a half point. The rule is not negotiable. ruclips.net/video/PFl6TxSA-SA/видео.html
@@warblerab2955 No. Sqrt (49) is always +7. Go check some arithmatic textbook. You are confusing it with the solutions to x^2 = 49. That is a different problem.
@@wernerviehhauser94 nope, sorry I think you are wrong. Sqrt of 49 has two answers, +7 and -7. Just take a look the link I post in here to another video on Sqrts done by tableclass math: ruclips.net/video/RXhIkfJ_Tf8/видео.html and this one as well: ruclips.net/video/kBaZYKhyBvQ/видео.html In both he indicates that square roots have two answer.
What is all this extra work? I did it in my head. Obviously if the sqrt of something plus the sqrt of something equals 2, then obviously 1 being the sqrt of 1 and 1+1=2, this means sqrt(5m+6)=1 and sqrt(3m+4)=1. Then 5m+6 = 3m+4. Then 2m=-2 and finally m=-1 then you go back and check it and it works.
"Obviously if the sqrt of something plus the sqrt of something equals 2, then obviously 1 being the sqrt of 1 and 1+1=2, this means sqrt(5m+6)=1 and sqrt(3m+4)=1"
This isn't correct. If you change "5m + 6" to "5m + 4", the solution becomes m=2*(4-sqrt(19)) and so it would have been wrong to conclude both square roots evaluate to 1.
However, it's a good heuristic to first assume that the square roots evaluate to integer values and see if that leads to an easy solution, because often in setting up algebra problems, that will be done to make the solution simpler stated.
Since √1+√1=2 I firstly thought that 5m+6 is equaled to 3m+4 so the answer is -1
Square root plus square root cancel , combine like terms than solve like a linear equation you get 8 ÷ -8 =-1
15 is NOT an extraneous solution. It is a real solution. sqrt(81) + sqrt(49) = 2 does work. Note that sqrt(49) = -7 is valid. (-7)^2 = 49, right? So sqrt(81) + sqrt(49) = 9 - 7 = 2.
Thank you Jakob, I was also wondering y he skated over the -ve root solution, not v rigorous for a maths teacher, I wouldn't employ him.
Great, clear video. For all the commenters saying that m=15 is also a solution, the symbol √ means the principal square root (ie, the positive square root). Yes, 49 has two roots, but the symbol √(49) means positive 7. That's why the graph of y=√(x) doesn't look like a full parabola on its side (just the "upper" half).
okay, maybe I am wrong. I always thought that if only the positive root was intended, a positive sign should be there to indicate such. Since neither a negative nor positive sign was shown (ei: y=-√(x) or y=-+√(x) ), I thought either root could be used. I thought that if only the positive root was intended, it should have said:+√(3m+4) and not just √(3m+4).
inventor of modulus function may rest in Piece #LOL
Mmmmh.
Following your assumption you can not write √49 = ±7, which is correct indeed.
I ever considered that y=√x describes a full parabola even if it is not a function because is not univoque.
@UC_P-LP8x5SDjnbDdKQfcdlA yes, is redundant but does not introduces additional solutions: {±√49} = {-7 , +7} = {√49} . BTW also the additional plus in +7 is redundant, but it is not wrong.
And because i can carrectly write √(3•15+4) = -7, I can also rewrite the initial equation substitutig that value to the second root:
√(5•15+6) + √(3•15+4) = 2
(±9)+(±7) = 2
At this point I must pick up the right solution among the values possibile. And if you catch +9 and -7 you have the equation verified.
Then m=15 is a solution.
The other values comes because the root is not a function.
If you assume that √49=-7 is wrong (which is not), then you can deduce that the only solution is -1. But your premise is false beacuse (-7)²=49, and so the conclusion is false too.
Really can you assert that the equation √(3x+4) = -7 has no solutions?
@@marcovoli Yup. But don't take my word for it, throw the equation into Wolfram Alpha or solve it with graphing software. And it's not what I'm saying, it's what the symbol √ is saying. That's the principal square root (ie, positive square root). It's a function. Does the equation x^2=49 have two roots? Yes! Does 49 have two square roots? Yes! Does the symbol √(49) represent both of them? No. It's true that we sometimes use the terms square root and principal square root loosely and interchangeably in common parlance, but the instant we write √ that ambiguity disappears. In the same way, arcsin(0.5) doesn't return every angle with a sine of 0.5, just the one that we, as humans, have defined as the most important. It's a sacrifice in completeness that we make for the sake of getting a helpful function. Same for the symbol √49. It doesn't represent every number that squares to 49 (or every solution to x^2=49), just the positive one.
Thanks! Forgot the multiple radical approach..
I quickly solved it by assuming integer values for the radicals so that sqrt(5m+6)=1 and sqrt(3m+4)=1 is the only possibility which gives 5m+6=1 and 3m+4=1 so m=-1. I know won't always work.
Very easy. Look:
a+b=2 and 3*a^2-5*b^2 = -2 . Easy quadratic equation.
Pretty straight forward.
I saw the equation, worked it out, then skipped to the end to see if I was right. I correctly got -1 and 15, but I didn't plug both in to see if they worked.
m = -1 looks like a simple solution. I haven’t watched your video yet, but the math works.
Obviously m has to equal -1 to get a solution that is an integer. Could square both sides and simplify, but that's a lot of work.
-1 may have been obvious, but it was possible that wasn't the only answer. That's why you do the extra work.
Great explenetion. Greetings from Poland
this is a Diophantine equation and the answer can be found by elementary analysis around m=0
No, this is not a Diophantine equation. For _two_ reasons: First, Diophantine equations are _polynomial_ equations - whereas the equation here includes roots. Second, for a Diophantine equation, one always specifies _beforehand_ that only integer solutions are sought - whereas here, solutions are sought over all the real numbers.
to all the commenters here saying it is -1. That is the answer, to the problem that is not solving the problem... Solve for the approximate value where the expression equals 3 and not 1....
both are correct solutions, because the answer to a square root is always 2 numbers +/-number
So you'd have 9 -7 = 2
instead of 9+7
Amazing how I draw blanks on some of these. More involved than I thought.
I got m = -1 just doing it in me' head. I just applied ^2 to every term.
What says you can't take the positive square root of the first and the negative square root of the second?
The square root of a non-negative real number is defined to be the non-negative number whose square gives the original number. So taking the negative value here would contradict the definition of "square root".
I could have solved it if it had been x instead of m. I'm not too good with equations involving m.
I went way off the deep end....I got the result:
4m - sqrt[quad. eq.] = -3
Pretty sure that's worse than I started!
As a year 10 sutdent in the UK (15 year old), I really struggle to understand how my american buddies find this type of question difficult. When i saw this and tried it it took only like 2 mins by squaring both sides, collecting, squaring again, collecting again and factorising the quadratic. Anyone else??? Any other UK friends???
Perhaps the answer to your question is genetic. Although I am California by birth, half of my DNA is of Britannic origin. I had no problem with the problem.
Get to the solution!!!!!! 3minutes +
My reasoning was exactly the same as Terry Flint, and since I have an HP RPN calculator under my fingertips, it took me approx 1 min to solve. However, I agree this would not be as simple in all cases. There were some give-aways here.
By any chance did you do it with the CAS on an HP Prime? The main advantage of the Prime CAS over that on the TI machines is that it does power series. (I'm still learning to use my WP-34s, which is not exactly an HP product but has an excellent solver.)
Nice way to flaunt😂
HELVET PACARDERS😅
Substitution: u^2 = 3m+4
i didnt know this thanks
I'm eating Crow. This is Way over My head.
Excellent service
m must be an integer, and it must be less than 0. Oh. It's -1.
Trivial :-)
Seemed clear cut when I first saw the problem. So, tried using plain old algebra. Could not solve it. Used an algebra app; still couldn't find it. Had to use graphing software. Found the -1 answer. Still, I can't follow the author's reasoning. Should have paid more attention to algebra in school.
This is a good one. Got my brain working quite a bit
You made this too hard.
Did you know that only 35% of the US Population possess Bachelor's Degrees? That is a fact. Wow! You're right about this too.
This is a great explanation of why tuition costs more than 500 times when I learned this stuff at UC Davis.
When I registered at Berkeley, my mother asked how much the tuition was. I told her, it was $67. She said, "That is outrageous; when I was there it was only $35 !"
Only glanced at it for a few seconds and could see the answer was -1, without needing any intermediate steps to simplify it. Probably trying to really solve it without simply seeing the answer would be more of a challenge. I think what helped was ruling out any positive number as yielding too large a result, while fractional values would have a hard time adding up to exactly one.
Same with me.
There was potentially more than one answer so that method doesn't necessarily come up with a complete solution.
@@hihoktf Yes, it was just somewhere between luck and intuition, not a formal solution technique. But intuition was also strongly saying that there would be either one solution or none. I recall a college course delving into the question of how you can try to determine whether an equation or set of equations has no solution, exactly one solution, or multiple solutions, but I don't recall all the details and need a refresher. There is a book by Kurt Godel, "On Formally Undecidable Propositions Of Principia Mathematica and Related Systems ", but so far it seems like such gobbledygook that I fear I will never understand either what is solvable or what is undecidable, so may have to rely on luck, intuition, or superstition, or simply keep trying things until I run out of ambition.
@@tom-kz9pb Godel was genius, but I take issue with self-reference (liar's paradox) in proofs. I think the idea of self-referential denial is incoherent. I do, however, ultimately believe all concepts only exist in contrast to other concepts so that all arguments at their core are circularly self-referential (ala chicken and egg). Tangentially related.. I'm not a fan of infinity (ala Cantor) either.
@@hihoktf I would have some philosophical issues with the very concept of "proof". What really is it supposed to mean? A line of argument so compelling that no person could disagree? As Devil's advocate, I hereby solemnly dispute every theory, proof, conjecture, axiom and postulate ever made or that ever will be made, thereby negating all attempts to define "proof" as a matter of universal agreement. Quite a profound thing for a mere RUclips comment. You could try to rescue "proof" by defining it as the thing that all serious experts must agree, but who gets to define who are the serious experts, and why should we buy it?
As for Infinity, I am on the other end of the spectrum. I think that we should have an "Infinity Appreciation Day" as a international holiday. Physicists usually don't like it when their equations start yielding infinite results, when probably it should be the other way around, a sign that they are on the right track. You would never get the magic of creation without genuine infinity at play. The universe is a product of random forces and could not achieve its creative miracles with mere googleplexes. Anything less than genuine infinity is insignificantly finite.
I squared all 3 sets at the start and ended up with -2/3. I was way off. Lol. Thank you for teaching me I can’t square 2 roots in the same expression
You were not way off. If you round to the nearest integer, you are right on.
Why acn't students solve problems in many situations? One reason is because they learn the wrong things. As in this video. At 3:40 the 1 "moves" to the rhs. I never told my students that we have operations like adding, substracting, multiplying but also an operation "moving". If I just move something it won't change. Why is it that the 1 becomes a -1 after I moved it. Why not stick tot real math? Just write about substracting 1 from both sides? Why again and again that strange way of dealing with operations in these videos? Just talk about what you really do: substracting the same value from both sides of an equation. That helps students to really realize what they do and helps them to make all kinds of fatal mistakes.
Me: f(x) = sqr(5x+6) + sqr(3x+4) is increasing so f(x)=2 has only one solution. And x=-1 is a trivial solution. The end...
Hated algebra. But love Trig and Calculus!
I wil describe my solution in portuguese, OK?
+++ Dá para se imaginar mentalmente que X ( ou m ) só pode ser menos UM
1) elevo ambos os lados ao quadrado
5X + 6 + 3X + 4 + 2 * ( RQ 5X + 6 ) * (RQ 3X + 4 ) = 4
8X + 6 = -2 * (15X² + 38X + 24) ( divido tudo por 2 )
4X + 3 = -15X² - 38X - 24 zero
then, -15X² - 42X - 27 = zero
X ( or m ) = - 1
That's OK, Folks!!!!!!
m= -1 was what I got after staring at the thumbnail for a couple minutes
Don't you simply square both sides of the original to get rid of radicals and collect like terms eg 8m =-8 and m=-1...or is this too simple???????
You have to square both sides of the equation. If you change something on the left side of the equal sign, you must make the exact same change on the right side in order to maintain equality. You squared the two radical expressions separately. You have to square the full expression on the left as well as the full expression on the right: [(sqrt 5m + 6 ) + (sqrt 3m + 4)] ^2 = 2^2 That is much harder to work with; thus, when you have two radical expressions on the same side of the equal sign, it's way simpler to put them on opposite sides of the equal sign and first eliminate one radical, then approach the other.
I am pretty sure you only squared the radixes and not the left side of the equation. If you did, you would end up with a product of radixes. Did you?
3:18 he gets to problem.
great problem. simple when you know you're gonna hafta isolate radicals twice
Agree - m = -1
we want the sum of the principle square roots to be 2. This can only happen if both square roots are equal to 1 or one root equals 0 and the other equals 2. The latter case gives the following systems
5m+6 = 4 or 3m+4=4
3m+4 = 0 5m+6=0.
Solving for m, it is clear these systems have no solutions, so we are left with the case that both principal roots are equal to one. This only happens when 5m+6=1 and 3m+4=1 , which implies m=-1. There is no need to do anything else, you are just making this confusing. Mathematics is about logic and using that logic as an art, not factoring crap repeatedly and following cookie cutter steps to get to some forced answer. Learn technique by approaching things in an unfamiliar way and it is usually much more elegant.
I suspect you're right. It seems much more clear cut the way you describe it. I should have noticed that intuitively.
Why did you assume the square roots had to be integers? It's a mistake to assume that. The video correctly verified all possible solutions. Your method arrived at an answer, but it was neither rigorous, nor, with all due respect, "elegant".
@@hihoktf I think you are quite correct.
@@hihoktf The case that one square root is an integer and the other is not can never happen. The case that either of them are irrational can never happen. So we are left with the case that they are both 1, the case that one square root is 0 and the other is 2, or the case they are both irreducible rational numbers. I explained everything except the last case (which is an axiom from elementary algebra).
"The sum of two irreducible rational numbers is an integer if and only if the divisors are equal. " in which a contradiction is formed and so we must be dealing with an integer m. This leaves only the case I presented which has solution m=-1.
For your arguments sake and the host of the video..... m could have been a complex number, a Quaternion, etc... The rationality of m is not specified in the question so integer can be assumed and if it is to be specified then elementary algebra (field theory) forces the argument into m being an integer as well.
@@nmayes1984 Perhaps I'm missing something from your argument. If m is a fraction, why wouldn't both 5m+6 and 3m+4 have the same divisors? I'm not seeing the contradiction you referred to there.
It just seems to me you didn't consider 5m+6=9/4 and 3m+4=1/4, or 5m+6=4/9 and 3m+4=16/9, etc. in your analysis.
And aren't 0, 1, and 2 irreducible rational numbers as is? Perhaps you meant "or the case they are both non-integer irreducible rational numbers"?
And yes, I had assumed a complex m, etc. was a plausibility.
What software are you using?
Is this click bait? You need to define the level of the student, I solved this in about 4 lines.
@LazyOldManAtHome: How old are you? If you solved this problem quickly you might be validating an important fact: older generations were far better at math than the current generation. This fact is the reason for the existence of this and other math channels--to fill in woeful gaps in math education left by our "woke" school system which is increasing good for nothing--except raising property taxes.
I looked at it as 2 - 10 = 8m, which means m/8 = -1. Am I on the wrong track? I think it comes down to the square root of 1 plus the square root of 1 = 2, but again I could easily be wrong.
I wrote the above without seeing the video, and since I used only general math, I thought my solution might be way off base. But now that I’ve seen the whole thing, I’m actually surprised that I was correct. :-) Here’s the way I approached it:
√(5m+6) +√(3m+4) = 2
√(8m+10) = 2
2 - 10 = √(8m
-8 = √(8m
-1 = √(1m
m = -1
√(-5+6) +√(-3+4) = 2
√(1) +√(1) = 2
@@MoosterMug I'm afraid to say you have numerous mistakes.
Step 2 does not follow from step 1 (you can't simply add them up the way you did) (if you could then sqrt(1)+sqrt(8)=3??)
Step 3 does not follow from step 2 (you can't simply move a number outside the square root) (if you could then sqrt(1+3)=4??)
Step 5 does not follow from step 4 (you wrongly divided the left side by one number (8) and the right side by a different number (sqrt(8)))
And step 6 does not follow from step 5 (sqrt(m) does not equal m unless by chance m=1)
You can test yourself by plugging m=-1 into every step of your proof. If your proof is good, that will result in no contradictions at any step. Unfortunately, your results are contradictory at every intermediate step. (i.e. step 2 would yield 1.414...=2??)
@@MoosterMug I'm afraid to say you have numerous mistakes.
Step 2 does not follow from step 1 (you can't simply add them up the way you did) (if you could then sqrt(1)+sqrt(8)=3??)
Step 3 does not follow from step 2 (you can't simply move a number outside the square root) (if you could then sqrt(1+3)=4??)
Step 5 does not follow from step 4 (you wrongly divided the left side by one number (8) and the right side by a different number (sqrt(8)))
And step 6 does not follow from step 5 (sqrt(m) does not equal m unless by chance m=1)
You can test yourself by plugging m=-1 into every step of your proof. If your proof is good, that will result in no contradictions at any step. Unfortunately, your results are contradictory at every intermediate step. (i.e. step 2 would yield 1.414...=2??)
@@hihoktf Thank you for the corrections. It appears that I arrived at the correct answer through pure luck. :-) But I hope you can see the logic even if it was completely off base. Going from step 5 to step 6 would never work with any numbers larger than 1, but in this instance, I believe my reasoning was sound if we discount all the steps above.
If -1 = √(1m, then m = -1. However, if -1 = √(2m, for example, then m would equal a different number. I believe that was your point.
@@MoosterMug Actually, assuming -1=sqrt(1m), the square root of -1 is i, not -1, so I think still a problem to say m=-1.
Math has changed since I was in school. Back when I had math class, the square root of 49 was either 7 or -7....
The square roots of 49 are the solutions to x^2 = 49 and therefore -7 and +7.
But the equation uses radixes, and the radix is by definition the principal (positive) square root of a real number. Using sqrt as calculator command is a poor choice of words causing the misunderstanding.
@@wernerviehhauser94 Like I said, things have changed since my day. We weren't taught to ignore half the solutions no matter how you wrote it down.
@@bentels5340 but that is exactly the nature of mathematical notation. It has exactly one defined meaning.
x^2 = 49 is not the same thing as x = √49.
So no, we are not ignoring solutions, there are no other solutions. If the negative value would be allowed, you could choose at every radix to use either the positive OR the negative value, hence you would have 4 possible solutions.
@@bentels5340 people knew how to do square roots thousands of years before negative numbers even existed, so originally it has always been like it is now. If anything it's your generation that attempted to change it.
Very interesting analysis but totally unnecessary! By looking at the equation it is just obvious that the value of m must be -1. So that’s how I did it . Square root of “5x-1+6” = -5+ 6= 1. The square root of “3x-1+4” = -3+4 = 1. 1+1=2. Too easy, and I never did math, history and English being my strong suits. Lovely test though!
It was absolutely necessary because there may have been more than one answer. To be strong in math, you must consider that possibility. To find a solution is not to guarantee a complete solution.
@@hihoktf Very interesting comment Doug. It seems that math is fluid. I had always thought that it was an exact science. When I first looked at the equation I thought that each part had to be 1, so that 1+1 =2, and that either part could not possibly be anything else, and if either part was different, then it would be an enormously difficult problem indeed. On that basis it was not hard to imagine that m just had to be -1, and it took abut 2 minutes to consider. I can see now that I did not do any math at all, and I was probably lucky that the equation was as it was! Thank you.
m=-1?
Convention 111010011,1101001(2)=
How do you know that “10 % of math students can solve this equation”? Did you collect and tabulate this from all of your students results?
Perhaps 🤔 you have access to a database of all the math students in the world?
oh boy... 90% won't have a clue on something which is fairly easy..... sad situation... :-(
I used the Distributive Property and got -1.
m= -1, just looking at the problem
It was much easier to brute force this problem than do the algebra, even checking to see if other answers were possible.
you might miss solutions
Thorin, what does brute force mean? Trial and error for all integers, real numbers? If the coefficients were tweaked slightly because this was a real life engineering problem, then what? Some sort of iterative technique? Possible, but now which approach is simpler and more accurate?
Nice 👍
Did you notice that if we subtract the radicals we get 2 as the solution?
Yes, but, by definition, sqrt(x) means the positive square root of x.
For example, x=sqrt(9) means x=3
In contrast,
x^2=9 means x=3 or -3.
M = 15 , M= -1
Far too complicated instructions.
I got m= 15 and -1
Literally solved it in my mind in 20 sec
Good but way too much going back, reviewing, sidetracks, etc (figured it out in my head in 2 seconds)