Solve and Check | Learn how to solve Rational equation quickly | Math Olympiad Training

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Can you solve the given rational equation for a? Step-by-step tutorial by PreMath.com
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Solve and check the rational equation
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Комментарии : 196

@ramanivenkata3161 Год назад

Excellent explanation

@PreMath Год назад

Excellent! Glad you think so! Thanks for your feedback! Cheers! You are awesome, Ramani. Keep it up 👍 Love and prayers from the USA! 😀

@雅玲陳-d8g Год назад

楊明傑 0:03

@popovoleg70 Год назад

@@PreMath hi from Russia. You just have forget memory function. Hence you can't program this and so it is art only. Memory is main thing, one of main things.

@davidbrisbane7206 Год назад

Let N = (a + 2)(a + 3)(a + 4)(a + 5). Let D = (a - 2)(a - 3)(a - 4)(a - 5). First we note that a = 2, 3, 4 or 5 are not solution, as then the demoninator is not defined. So, these values are excluded from the analysis below. Now we note that N/D = 1, if and only if (N + 1)/(D + 1) = 1. This is clear because if N/D = 1, then N = D, and so N + 1 = D + 1, so (N + 1)/(D + 1) = 1. Similarly, if (N + 1)/(D + 1) = 1, then N + 1 = D + 1, so N = D, and so N/D = 1. Now, N + 1 = (a + 2)(a + 3)(a + 4)(a + 5) + 1 = (a + 2)(a + 5)(a + 3)(a + 4) + 1= (a² + 7a + 10)(a + 7a + 12) + 1 = (a² + 7a + 10)(a² + 7a + 10 + 2) + 1 = (a² + 7a + 10)² + 2(a² + 7a + 10) + 1 = (a² + 7a + 11)². Similarly, D + 1 = (a² - 7a + 11)² Hence (N + 1)/(D + 1) = (a² + 7a + 11)² / (a² - 7a + 11)² If a > 0, then (N + 1)/(D + 1) > 1. So N/D > 1, which contradicts the given condition that N/D = 1. If a < 0, then (N + 1)/(D + 1) < 1. So N/D < 1, which contradicts the given condition that N/D = 1. If a = 0, then by direct computation, N/D = 1. Hence, a = 0 is the only real solution to N/D = 1.

@wesleysuen4140 Год назад

[A graphical approach] Let f(x)=(x+2)(x+3)(x+4)(x+5). Then we are to solve f(a)=f(-a) for real values of a. Consider the graph G of y=f(x). It is a W-shaped polynomial curve intersecting the coordinate axes at (-5,0), (-4,0), (-3,0), (-2,0) and (0,120). It is also symmetric about the line x=-7/2 which intersects the curve at its local maximum point (-7/2,9/16). Now also consider the graph H of y=f(-x) which is the reflection image of that of y=f(x) in the y-axis. By the reflection symmetries as discussed above, the left arms of the graphs G and H go parallel to each other, and so do the right arms, with no intersections. And since the y-intercept 120 is (a lot) greater than the local maximum 9/16, the graphs G and H intersect cannot intersect with each other any more other than at x=0. That is to say, a=0 is the only real solution to our problem.

@misterenter-iz7rz Год назад

0 trivially.😃

@sa4dus Год назад

you must proof there are no more real solutions

@misterenter-iz7rz Год назад

@@sa4dus I just means 0 is clearly one solution, and I do not means no other solution may exists. Actually premath show that this does not exists.😄

@PADABOUM Год назад

@@sa4dus trivial as well. Numerator and denominator need to be equal, hence there absolute value need to be equal. If a>0 then then l a+n l > | a -n l and hence any product of such is also > to the product of la-nl Is a< 0 then the reverse is true Hence 0 is the only solution that is real

@josephwilles29 Год назад

Thanks for nothing

@PADABOUM Год назад

@@josephwilles29 why? Isn’t my explanation clear?

@luxel3607 Год назад

Great video! But there is no way this is a math Olympiad question, is far too easy. What Math olympiad did the question appeared on?

@haisonnguyen1573 Год назад

x=a^2+7a+11, y=a^2-7a+11 --> (x-1)(x+1)=(y-1)(y+1) --> x^2-1=y^2-1 --> x=y or x=-y --> a^2+7a+11=a^2-7a+11 or a^2+7a+11=-(a^2-7a+11) --> 7a=-7a or 2a^2+22=0 --> a=0

@arthur_p_dent Год назад

Multiply the equation by the demoninator and expand, result is a^4+14a³+72a²+154a+120 = a^4 - 14a³ + 72a² - 154a + 120 Simplifying, we get 28a³ + 308a = 0. Factor out the zero, and the rest will simplify to a²=-11. Thus, the solutions are 0 and +/- sqrt(11) *i. one has yet to check whether any of these solutions will produce a zero in the denominator of the original equation. This isn't the case, therefore all 3 of these are valid solutions.

@plrc4593 Год назад

Seems much simpler :D

@dongxuli9682 Год назад

no need to calculate even power of a terms, since all even powers terms have the same sign on both sides. Only odd terms left, and all coefficients are positive, so only real solution is 0.

@arthur_p_dent Год назад

@@dongxuli9682 yes, 0 is the only real solution. But we want the complex solutions as well.

@cvb-bm5dg 12 дней назад

Well done!

@nineko Год назад

I know that you had to reject the complex solutions because you were looking only for the real ones, but when you have a² = -11, it should result into a = ±√-11, while you only listed the positive case. It's a minor nitpick, but I thought to mention it for completeness.

@PreMath Год назад

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@AdityaKumar-gv4dj Год назад

@dilipbokil5531 Год назад

It is given that 'a' is a real number. otherwise you are also right

@VagishaDas Год назад

0 should work because 0-# and 0+# has an exact opposite value and as the items in the multiplication are even you will always get a positive number. Any number divided by itself is 1.

@jollyjoker6340 Год назад

And any negative number makes the denominator larger and the numerator smaller (well, the numerator starts rising again after a while but never to equal the denominator) and vice versa, so 0 is the only solution. I assume you can show this with some simple inequalities

@MrSeezero Год назад

First I figured that I could put the denominator to the right side to have two 4th-degree polynomials set equal to each other. Using Vieta's rule on how roots and coefficients are related in a general polynomial, I was able to eliminate the 4th-, 2nd-, and 0th-degree terms from both sides of the equation to get 14a^3 + 154a = -14a^3 - 154a. I then went 14a(a^2 + 11) = 14a(-a^2 - 11); 14a(2a^2 + 22) = 0; a(a^2 + 11); a = 0, +sqrt(11)i, -sqrt(11)i; a = 0 only after only considering the real solution(s).

@sudiptaraxit3778 Год назад

a=0 is immediately observed. Without using pen and paper.

@deusexmachine2834 Год назад

Halo Jamal thank you for wis vedea ai hope you mak mor contanta in tha futur

@PADABOUM Год назад

As the question was asking to solve quickly, this would not be the best way. Fastest way is to understand denominator and numerator need to be equal, hence their absolute value need to be equal, for a positive absolute value of a+n > absolute value of a-n and for a absolute value of a+n. Multiply those and the order stays the same, hence 0 is the only answer.

@АнатолийИванов-й1ъ Год назад

а = 0 сразу бросилось в глаза. Перемножить 4 минуса - будет плюс, как и в числитии, а,, а " только мешает. Как - то так.

@ho-jinlee2468 Год назад

Let f(x) be the numerator, (x+2)(x+3)(x+4)(x+5) and g(x) be the denominator, (x-2)(x-3)(x-4)(x-5). Since f(-x) = g(x), it means they are symmetric against the y- axis. Observe that they are both 4th degree functions with positive coefficients of 4th power terms, the largest x value for which f(x) =0 is -2 and the smallest x value for which g(x)=0 is 2. It implies that the common x value for which f(x)=g(x) is 0 only, ie, they only meet at x=0.

@ピヨちゃん-r7b Год назад

(a+2)(a+3)(a+4)(a+5)=(a-2)(a-3)(a-4)(a-5) af(a)+120=ag(a)+120 and F(x)=(x+2)(x+3)(x+4)(x+5), G(x)=(x-2)(x-3)(x-4)(x-5) and the gragh of y=F(x) is a form of W(when 0

@Officially_Kimdeheizer Год назад

Approach 2: Do cross multiplication first. There are four terms so as per the conditions A=0, will satisfy the solution as per signs. As only A is the only variable thus A is only equal to 0.

@g.r.onlineclass Год назад

Nice 👍🙂👍

@PreMath Год назад

Thank you! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@nasrullahhusnan2289 Год назад

The numerator of your first hand writing may be written as [(a²+7a+11)-1[][[a²+7a+11)+1] simplified into (a²+7a+11)²-1 In similar way the denominator becomes (a²-7a+11)²-1 Hence (a²+7a+11)²-1=(a²-7a+11)²-1 (a²+7a+11)²-(a²-7a+11)²=0 [(a²+7a+11)+(a²-7a+11)][(a²+7a+11)-(a²-7a+11)=0 2(a²+11)14a=0 --> a=0 a²+11=0 --> a=±i×sqrt(11)

@davidbrisbane7206 Год назад

Given (a + 2)(a + 3)(a + 4)(a + 5)/(a - 2)(a - 3)(a - 4)(a - 5) = 1. Then clearing the demoninator and placing all the terms on the LHS of the equation, we obtain (a + 2)(a + 3)(a + 4)(a + 5) - (a - 2)(a - 3)(a - 4)(a - 5) = 0. This simplifies to 28a³ + 308a = 0. So a(28a² + 308a) = 0, which has solutions a = 0, a = i√11 and a = -i√11. However, if we use Descartes Rule of Signs on 28a³ + 308a = 0 to look for real roots of this equation, we find it has no positive real roots and no negative real roots. But it must have at least *one* real root as it is a cubic equation, so the only possibility for this root is a = 0, and of course it is a root, and it is indeed the *only* real root.

@musicsubicandcebu1774 Год назад

I wouldn't call that a 'quick' solution though - more like brute force. Thanks anyway, great job.

@peterlim8416 Год назад

A quick look at the +++ above, and --- as divider, the resulting number can only expanding, but it's 1 instead. This does not make sense. Because the group is in even number, so only a=0 going to be possible. This can be determined logically.

@ashwanibeohar8172 Год назад

By simple logic a cant be 1,2,3,4,5 or -1,-2,-3, -4,-5 as it makes one factor zero, which results in soln as either zero or infinite. And for 'a ग्रेटर then 5' sol will be more than 1, and for 'a less then -5' sol will become less than 1. Only a=0, satisfy the eqn.

@batavuskoga Год назад

I did it differently with the same result : a=0 (a+2)(a+3)(a+4)(a+5)=(a-2)(a-3)(a-4)(a-5) After calculation you get a^4+14a³+71a²+154a+120=a^4-14a³+71a²-154a+120 14a³+154a=-14a³-154a 14a³+14a³+154a+154a=0 28a³+308=0 a(28a²+308)=0 --> a=0 28a²+308 : a=sqrt(-11) --> not a real solution

@davidbrisbane7206 Год назад

Small correction 28a³ + 308a = 0. But good solution 👍.

@batavuskoga Год назад

@@davidbrisbane7206 That's a mistake in typing. On paper I did it correct

@davidbrisbane7206 Год назад

@@batavuskoga No problems 👍

@sudhanshu1709 Год назад

I swear to God I solved this one in 5 seconds

@erwinsoetens1604 Год назад

very nice explanation , thx

@edkerry87 Год назад

This is a prototypical MO substitution/pattern problem. try -1, 0 and 1 and you will immediately see that as increases the quotient increases beyond 1. As decreases, the quotient decreases below 1 approaching 0. 0 is the only real number that works.

@drrageshprathapan7289 Год назад

I have solved this problem in a different way and in few steps. As per the question, numerator and denominator of the fraction has to be equal to make it 1. So we can iterate values starting from a=0, 1,2, etc. and check whether both numerator and denominator is equal. At first step itself the value of a=0 was found to be appropriate for this problem, hence a=0. PS: you can try with other numbers but the check for equality between numerator and denominator will fail.

@ActMedInfo Год назад

А я тупо раскрыл скобки, в результате получил a(25a^2+29a+308)=0; Или a=0 - т.е. ответ задачи, или 25a^2+29a+308=0. Вольфрам дал такое решение: a = -29/50 ± i sqrt(29959)/50 Проверять я это, конечно, не буду)

@josukehigashikata1494 Год назад

By observation a = 0.

@lurefishing5350 Год назад

Since a is a real number, it is sufficient to consider a real number space figure. 1) y=f(a)=(a+2)(a+3)(a+4)(a+5) 2) y=g(a)=(a-2)(a-3)(a-4)(a-5) f(a) : y=0 for a=-2, -3, -4 and -5. g(a) : y=0 for a=-2, 3, 4 and 5. Both f(a) and g(a) are downward convex quadratic functions. Also, f(a) and g(a) are symmetric along the y-axis, so g(a) > f(a) in the region a5. Therefore, f(a) = g(a) is satisfied only for a=0.

@Psykolord1989 Год назад

Before watching: Only real solutions? Alright then, a=0. Numerator is now 2*3*4*5, denominator is (-2)(-3)(-4)(-5). We have 4 negatives on the bottom, meaning they will cancel out, ans thus denominator is essentially (2*3*4*5). You then have (5*4*3*2)/(-5*-4*-3*-2) = 120/120 = 1.

@ВалерийЦелищев-ш6з Год назад

a # 2, 3, 4, 5 => (a + 2)(a + 3)(a + 4)(a + 5) = (a - 2)(a - 3)(a - 4)(a - 5) I'm lazy enough to multiply 4 brackets, but I can multiply 2!!! (a^2 + 5a + 6)(a^2 + 9a + 20)=(a^2 - 5a + 6)(a^2 - 9a + 20) Let x = a^2 + 6, y = a^2 + 20 => (x + 5a)(y + 9a)=(x - 5a)(y - 9a) Oh!!! I can do it!!! xy + a(9x + 5y) + 45a^2 = xy - a(9x + 5y) + 45a^2 However, there is something the same in both parts => a(9x + 5y) = -a(9x + 5y) => 2a(9x + 5y) = 0 => a = 0 & 9x + 5y = 0 I still remember what x & y equals and after a couple of manipulations: 14a^2 + 154 = 0 or a^2 + 11=0 :( Only comlex solution.. Therefore a = 0 only... Sorry for the empty lines, I think it's easier to read this way.

@larisamedovaya9097 Год назад

Consider denominator: “a” can ‘t be equil 2; 3; 4;5 be cause denominator can’’t be =0; consider signs of denominator : number of signs =4, So result always positive; any value of “a” is not solution because numerator and denominator will be different, but a=0 is our solution and what we are looking for. Check ( (0+2)(0+3)(0+4)(0+5))/((0-2)(0-3)(0-4)(0-5))=120/120=1 GED

@337호끼리 Год назад

a is real so, Let every number (a-5, a-4, ... , a+4, a+5) placed on number line. Then each number a-5, a-4, a-3, a-2 0. And also -(a-2)=a+2, -(a-3)=a+3, -(a-4)=a=4, -(a-5)=a+5 Therefore a is 0

@user-u3e3kx5m6pg Год назад

If a>0, |a+k| > |a-k| and if a |a+k| for k= 2,3,4,5. So, in both cases absolute value of (a+2)(a+3)(a+4)(a+5)/(a-2)(a-3)(a-4)(a-5) is not 1. Thus only possible number a is 0 which works

@davidbrisbane7206 Год назад

Another less (computational) algebraic way of solving the equation for a, where a is a real number. Let N = (a + 2)(a + 3)(a + 4)(a + 5) and Let D = (a - 2)(a - 3)(a - 4)(a - 5) and given N/D = 1. First we note that a = 2, 3, 4 or 5 are not solution, as then the demoninator is not defined. So, these values are excluded from the analysis below. *Case (a > 0)* Now a + 2 > 0 a + 2 > a - 2 and a + 3 > a - 3 and a + 4 > a - 4 and a + 5 > a - 5. So, N/D ≠ 1, which is contrary to the given N/D = 1 So, there are no solutions to the equation where a > 0. *Case (a < -0)* Let u = -a, where a < 0 So u > 0,. So, N = (-u + 2)(-u + 3)(-u + 4)(-u + 5) = (u - 2)(u - 3)(u - 4)(u - 5) So, D = (-u - 2)(-u - 3)(-u - 4)(-u - 5) = (u + 2)(u + 3)(u + 4)(u + 5) So, N/D = (u - 2)(u - 3)(u - 4)(u - 5)/(u + 2)(u + 3)(u + 4)(u + 5). Now u + 2 > 0 u + 2 > u - 2 and u + 3 > u - 3 and u + 4 > u - 4 and u + 5 > u - 5. So, N/D ≠ 1, which is contrary to the given N/D = 1 So, there are no solutions to the equation where a < 0. *Case (a = 0)* N/D = 2*3*4*5 / [(-2)*(-3)*(-4)*(-5)] = 1. So, a = 0 is a solution to the equation N/D = 1. As we have covered all possible real value cases, we conclude that a = 0 is the only solution to N/D = 1, where a is real.

@ايمنكويكي Год назад

In the denominator ,it should a is different 0so a=0 in the nominator

@PreMath Год назад

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@thefuture2411 Год назад

có A +Z > A-Z với mọi A thuộc R và Z>0, => (A+Z)^n/ (A-Z)^ n khác 1 với mọi n là số lẻ, xét (A+Z)^2n/ (A-Z)^2n =1 ^n => 2AZ= -2AZ => A=0

@yoops66 Год назад

Sorry but I don't like when a solution comes by arranging things without any smart reason, and that, magically, things get simplified. You learn nothing doing that. Here, what was important is that by doint one multiplication, in any combination, you're ending with a²+2ab+b² on the left side and a²-2ab+b² on the right one. You just need to do this twice and you may predict that a lot of things will simplify (at least a^4 and the independant term, maybe more). If I'm not wrong, doing the extensinve calculation- leaves indeed -2a³=2a³, thus a=0. Maybe is it more brute force, but not that much because you know that things will simplify, and you're sure you may reproduce such thing in other exercices. Onthe other hand, magical resolution can usually not be generalized.

@giuseppemalaguti435 Год назад

a=0,a^2=-11

@PreMath Год назад

We are only looking at real numbers!

@enalaxable Год назад

a=15.666360... Non rational real solution of the 3rd deegre polynomial, after removing the a factor. How can one miis that? I mean a 3rd degree poly ALWAYS has at least a real solution.

@AjayYadav-jc7hw Год назад

my answer is( -11)½ is correct or incorrect

@dipeshpatel1934 Год назад

As stated in the video, *a* is a real number.

@PreMath Год назад

Dear Ajay, We are only looking at real numbers! Cheers

@KAvi_YA666 Год назад

Thanks for video.Good luck sir!!!!!!!!!!!!

@PreMath Год назад

So nice of you, dear Thanks for your continued love and support! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@lu.cicerone.cavalheiro Год назад

What truly baffles me is why hasn't he multiplied both sides of original equations by (a(a+1))/(a(a-1)) to create two arithmetical progressions on left side... it would be faster.

@danishm.3417 Год назад

You don't understand maths. That was a show of rot learning, not a systematic approach. What's the lesson for the students? What fundamental technique should they understand and remember? Such a fundamental approach can only be of axiomatic nature. Since you lack that understanding, a well-deserved dislike!

@harrymattah418 Год назад

If a>0, then each term at the numerator has an absolute value greater than the absolute value of its counterpart at the denominator. For instance |a+3|>|a-3|. This quotient can never be 1. If a

@TonyHammitt Год назад

Based on the thumbnail (no "a has to be real" qualifier), I saw 0 as a solution and guessed about 3.5i as another solution. Not too far off 😀

@Hopeforgooddays Год назад

I am very happy to see it, that's you are very talented and knowledgeable person ,which comments at my videos and likes my videos. Thanks alot for your kind support,for your precious time ❤️,for your prayers. I always pray for you. You get success and happiness in every step of life. Aameen Aameen

@PreMath Год назад

Many many thanks So nice of you. You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Take care...

@nitukumari4084 Год назад

First view😆😁

@PreMath Год назад

Good job! Thank you! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@picturetaker607 Год назад

This problem can benefit from observation and a bit of logic before diving in and solving. First, the equation equals 1, which means the numerator and denominator must equal each other. Second, the denominator can not equal zero, therefore a can not be 2,3,4,5. and by inspection any integer for a is not going to solve the problem. try zero and it works.

@sr6424 Год назад

Logic gave me the real solution quickly. The numerator and denominator must be equal. Hence multiplying 4 negatives will give a positive. I’d like to see you prove the complex solutions!

@bird-honey Год назад

08:50, I don't know why the formula below works like that. (x-y)(x+y) + 2(x-y) => (x-y)(x+y+2)

@crir85 Год назад

What I did was this: separated equasion in 4 terms (a+2)/(a-2) * (a+3)/(a-3)*... thought that for this to equal 1 I must have 1 or -1 for all four terms. Solution was clear: a=0.

@weggquiz Год назад

good job, thanks for a nice video

@PreMath Год назад

Thanks for watching! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@MsRa3d Год назад

I solve it in few seconds 1st a is not equal 2 3 4 5 denominator should not equal zero. 2nd a should be great than 6 or less than 2 for + denominator. 3rd lit a +inf = +infini 4th limit a -inf = 0+ the only solution is 0.

@subliminalfalllenangel2108 Год назад

0. Anyone here solved equations like this without having to write it down? Edit: okay, I didn't figure out the imagined numbers.

@sudiptaraxit3778 Год назад

It's immediate to notice that the constant term and the 4th order term cancel out. So on both sides you would have cubics with zero constant. So a=0 is a solution...do not need to use even a pen. This cannot be a Math Olympia problem. Too easy.

@gauss17 Год назад

We are only looking for real solutions. For a>o the numerator is clearly larger in absolute value than the denominator. For a

@DB-lg5sq Год назад

شكرا لكم ليكن a يخالف 2,3,4,5 نضع A/B=1 في حالة a=0,فإنA=B في حالة aيخالف0فانA>B اذن AيخالفB نجد الحل الوحيد هو a=0

@huyminhha658 Год назад

i rate this solution 8/10 because you forgot to find the definite condition of a (a#2;a#3;a#4;a#5)

@HappyFamilyOnline Год назад

Great👍 Thanks for sharing🌺🌺

@Hopeforgooddays Год назад

Your all videos are very helpful and nice for students. Always stay blessed and stay connected plz.

@PreMath Год назад

Thank you, I will! You are so kind. Keep it up 👍 Love and prayers from the USA! 😀

@zhongxichao1976 Год назад

Technically, as it is a fourth order equation, there should be 4 solutions? You should also give the other 3, or prove that alll 4 solutions are 0.

@bbloodjh Год назад

Using symmetry, a should be zero because letting a=-X gives rise to the exactly same equation.

@Mine_Is Год назад

@cosimo7770 Год назад

In 2 seconds we see that a=0 is a solution. 4 negatives multiplied in the denominator gives a positive. So numerator and denominator are equal i.e. 1

@antoinegrassi3796 Год назад

Pardonne-moi, je suis toujours aussi fana de tes vidéos, mais tu as oublié la première étape NÉCESSAIRE avant de commencer une quelconque étude sur une équation. On doit calculer son DOMAINE de DÉFINITION, et l'ensemble sur lequel on l'étudie. Et ceci même si ce domaine est souvent implicite ou évident. Donc ici on commence l'étude en indiquant qu'elle se fait sur R \{ 2; 3; 4; 5} et on la termine en verifiant que les solutions trouvées appartiennent bien au domaine de définition. Merci de ne pas considérer ceci comme du superflu, mais comme une rigueur nécessaire. Remarque: le symbole " \ " se nomme "anti-slash" et peut se lire de plusieurs façons: " R moins { 2; 3; 4; 5}" ou " R privé de..." ou " R sauf...". Le "moins" étant compris au sens de la soustraction d'un ensemble, c'est à dire ici de " l'intersection de R avec le complémentaire de { 2; 3; 4; 5} dans R" . Comme c'est fastidieux, il faut le traiter avec légèreté et simplicité. Un exemple: résoudre sur R l'équation: rac(-1 - x^2) = 4. Sur R, le domaine de définition est l'ensemble VIDE . STOP. L'étude est terminée. L'ensemble des solutions est VIDE lui aussi. 😉 Salut

@figerdron_8972 Год назад

Can it be solved by using arithmetic progression? x=a+2; x+1;x+2..

@KRYPTOS_K5 Год назад

Could I get the product of the numerator by some formulae (which are the 4 terms of an arithmetic progression with r=1 and n=4) and equalize it with the denominator which is also the product of the terms of an arithmetic progression (n=4, r=-1)?

@davidbrisbane7206 Год назад

Let N = (a + 2)(a + 3)(a + 4)(a + 5). Let D = (a - 2)(a - 3)(a - 4)(a - 5). Now we note that N/D = 1, if and only if (N + 1)/(D + 1) = 1. This is clear because if N/D = 1, then N = D, and so N + 1 = D + 1, so (N + 1)/(D + 1) = 1. Similarly, if (N + 1)/(D + 1) = 1, then N + 1 = D + 1, so N = D, and so N/D = 1. Now, N + 1 = (a + 2)(a + 3)(a + 4)(a + 5) + 1 = (a + 2)(a + 5)(a + 3)(a + 4) + 1= (a² + 7a + 10)(a + 7a + 12) + 1 = (a² + 7a + 10)(a² + 7a + 10 + 2) + 1 = (a² + 7a + 10)² + 2(a² + 7a + 10) + 1 = (a² + 7a + 11)². Similarly, D + 1 = (a² - 7a + 11)² Hence (N + 1)/(D + 1) = (a² + 7a + 11)² / (a² - 7a + 11)² If a > 0, then (N + 1)/(D + 1) > 1. So N/D > 1, which contradicts the given condition that N/D = 1. If a < 0, then (N + 1)/(D + 1) < 1. So N/D < 1, which contradicts the given condition that N/D = 1. If a = 0, then by direct computation, N/D = 1. Hence, a = 0 is the only real solution to N/D = 1.

@khalidaljuhani-ut7jx Год назад

you have a big mistake in your solution that is X&Y as per my joiner knowledge you cant take a part of bracket

@idzhaman Год назад

It was not quickly. I did it much faster by just multiplying without tricks.

@MuhammadSaleem-hj5ej Год назад

Why u r using this long method, directly put a equal to zero in n and d for sharp calculation.

@amarendrabhowmick8896 Год назад

Good but since a is a real no. a^2+22 =0 is not possible .no need of extending to complex set.

@ХекфиВол Год назад

0 разумеется! 4 множителя, чётное число, значит знак одинаковый.

@bertblankenstein3738 Год назад

a=0 by inspection, but then the thorough steps of proving there are no more real solutions.

@williamestes629 Год назад

Not exciting, but I've forgotten how to do so much math like this.

@shubhamjaiswal192 Год назад

I have solved it by just thumbnail, without pen and paper.

@joseriera3638 Год назад

a > 0 => numerador > denominador, a < 0 => numerator < denominator, hence a has to be =0

@북극곰-u8z Год назад

By symmetry & parallel shifting, the solution is x=0 solely

@Adventurin_hobbit Год назад

Bro haven't you heard about a concept called ratios and proportions

@valeriykotlov2763 Год назад

The worst possible solution written by not a mathematician.

@supercuber1915 Год назад

Explantion okay but your voice doesn't fit me

@evbdevy352 Год назад

Thanks s lot for your explanation.Good luck.

@ColonelFredPuntridge Год назад

A=0, obviously. (That took about one half of a second.)

@алексейслобожанинов-л4ю Год назад

Сколько блуда! С первой прикидки понятно. Ставь "0".

@ivantochev3980 Год назад

(a+2)/(a-2) must be 1 or -1. So we have 4x (-1) which is 1.

@am2564 Год назад

Ur math is the amazingly amazing!🎉

@СтаниславАнтонов-э7ч Год назад

Решается за пару секунд без всяких вычислений

@mauijttewaal Год назад

Much Too Complicated explanation...

@ashwanibeohar8172 Год назад

For all positive वैल्यूज of "a", all factors of denominator become less than neunator thus the डिवीज़न become more than one. And for all negative values of a, Denominator become positive and all four positive factors of denominator become more than the four factors of neumnator, thus the devision become less than 1. Only a=0 can satisfy the eqn.

@keithwood6459 Год назад

Yes, I saw this too. It converges on 0 being the only possible answer in the real numbers.

@ganeshdas3174 Год назад

Go for a= 0 for equality of this rational number

@jim2376 Год назад

By inspection a = 0 works. 120/120 = 1.

@terrellcampbell2875 Год назад

0. Figured that out pretty quickly.

@abcxyz-nd6xh Год назад

Ｚ　Ｅ　Ｒ　Ｏ in 10 sec 4 (Even No.) of Minus Below => Donominator is 　╬ＶＥ＝＞　Ａ　Ｍｕｓｔ　Ｂｅ　ＮＯＴＨＩＮＧ　　　　ＴＯ　ＭＡＫＥ　ＵＰＰＥＲ　＝　ＬＯＷＥＲ

@ganeshdas3174 Год назад

On expanding numerator left (a^2+7a+10)(a^2+7a+12)= a^4+a^3(7+7)+a^2(10+12+49)+a(84+70+120 =a^4+14a^3+71a^2+154a+120 Similar expansion will come with alternate+ve and -ve signs on right hand side also .Cancel similar sign terms from both the sides. Leaving 14a^3+154a = -14a^3-154a 28a^3+308a=0 or,28a(a^2+11)=0 As such ,a =0 or a = + - ✓11i

@thomakondaciu6417 Год назад

a=0.a = rrenja katrore e minus 11

@vasileioskosmidis1421 Год назад

it is so simply and you make it so complex.

@priyanthasubasinghe931 Год назад

Great explain Thank you sir

@alster724 Год назад

Got the drift. Like the last video, tricky at first but was able to catch the concept

@pavlozemskyy4935 Год назад

if a not equal to 0, the sentence above is always greater than sentence below. The only point where sentence above can be equal to the sentence below is if a = 0. Simple check and we are done.

@PADABOUM Год назад

No, if a is negative, it is the other way around but the result 8s the same.