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A couple of innacuracies on the solution: 1. a^2-b^2 is not a perfect square, at least not in general. It is the difference of squares, and sure, there is a nice factorization formula, the one you used. 2. The product 4x2x22 is 176, not 88.
Знаменатель переносим вправо. Слева и справа от знака равенства полиномы имеют симметричные корни относительно 0, значит надо проверить X=0. Это корень, других корней не может быть, так как увеличение одной строны равенства приводит к уменьшению другой
The product of the distances from point X to 2, 3, 4 and 5, and The products of the distances from point X to -2, -3, -4 and -5 are equal, The location of the X point at this time is a problem. Since 2, 3, 4, and 5 and -2, -3, -4, and -5 are symmetrical to each other around 0 in the coordinates, The position of point X is 0.
For any positive value of x, the numerator will be a larger number than the denominator. And for any negative value of x, the numerator will be a smaller number than the denominator. Both cases make it impossible for the quotient to be equal to 1, therefore x = 0 is the only possible answer.
I just had an interesting solution in mind, in this ques we are solving (x-2)(x-3)(x-4)(x-5) = (x+2)(x+3)(x+4)(x+5). When we plot the graph for left hand side and right hand side, we will find 2 w shape curves in which - 5,-4,-3,-2 & 2,3,4,5 are zero. These two curves obviously will only have one intersection Point. And as we know the two curve are symmetric, so the intersection occurs in x=0.
Nicely done. I don't think you need to appeal to the discriminant though. If 2x^2 + 22 =0 it is fairly obvious that x^2 is negative, but I do see the appeal of completeness.
Bring the denominator to the other side. Expand. The fourth, second and zeroth powers of "x" fall away. Left with an equation in x and x cubed. One root is zero, left with an equation: x squared equals constant.
If you won to teach math atlist you have to know maths. 4X2X22=176 and not 88 Second. For this kind of problem the first is that all this in denominator have to be different from zero. That is requirement. So please.....
Cross multiply. Distribute. The forth powers and the constant terms cancel each other. Put everything on the same side. Remains a cubic that equals 0. Factor out x. x=0 is a solution. Remains a quadratic with two complex solutions.
As the numerator and denominator have four factors, each of which is similar differing only in the sign (the factor in the numerator is a sum but in the denominator it is a difference) it easily seen that when x=0 they are the same in number but different in sign (plus in the numerator and minus in the denominator). Their product will be the same both numerical value as well as sign. The ratio is 1.
Ну оно не единственное, о комплексных корнях мы не забываем, однако для этой задачи это верно, так как в этом видео в условии сказано, что Х принадлежит только множеству действительных чисел
Автор-зануда, расписал фигню на 12 минут.Мне, историку по образованию, окончившему советскую школу ровно полтос лет тому назад очевидно, что неизвестное нуль.Вспоминается, когда занимался матекой с племянником, быстро говоришь, если равное разделить на равное, получишь...бедный ребёнок успевает выпалить (получишь) равное.Я менторским тоном : нет, получишь единицу.
So simple to simply infer equality of numerator and denominator, of course 0 is one solution because applying it just leaves a positive number each side, but then for any other X it is an impossibility...
At this point 4:09 you can make the solution much faster and easier by set a= x2+7x+11 and b=x2-7x+11 instead. Then we have (a-1)(a+1)=(b-1)(b+1), means a2-1=b2-1 >> a2=b2 >> a=b or a=-b.
Here is my solution (of course way faster): 1) We work on the numerator. If we introduce y=x+7/2 our expression becomes: (y-3/2)(y-1/2)(y+1/2)(y+3/2) Of course (y-1/2)(y+1/2)=y²-1/4 and (y-3/2)(y+3/2)=y²-9/4. Now let's transform (y²-1/4)(y²-9/4). With the same idea we introduce z=y²-5/4. Our expression becomes: (z-2)(z+2)=z²-4. So the numerator can be written as (y²-5/4)²-4 or as [(x+7/2)²-5/4]²-4 2) The same work on the numerator would lead to the expression [(x-7/2)²-5/4]²-4 3) Our problem is equivalent to solve (for x not equal to 2, 3, 4 or 5): [(x+7/2)²-5/4]²-4=[(x-7/2)²-5/4]²-4 or [(x+7/2)²-5/4]²=[(x-7/2)²-5/4]² We have two cases: a) (x+7/2)²-5/4=(x-7/2)²-5/4 or (x+7/2)²=(x-7/2)² Since x+7/2 can't be equal to x-7/2 this implies that x+7/2=-(x-7/2)=-x+7/2 then x=-x then x=0. b) (x+7/2)²-5/4=-[(x-7/2)²-5/4] or (x+7/2)²+(x-7/2)²=5/2. We can see that among x+7/2 and x-7/2, at least one is outside [-5/2,5/2] then its square is >25/4>5/2 and same for the sum. So we know that we don't have any real solutions. We can still find the complex solutions: 2x²+49/2=5/2 => x²=-11 then x=sqrt(11).i or -sqrrt(11).i
Sorry, but at the determinant, in what universe does - (4 x 2 x 22) =- 88? - 4 x 2 = - 8. - 8 x 22 = - 176. And, I really don't get why that's needed anyway as the substitution result of 2x² + 22 = 0 leads to the much easier 2(x² + 11) = 0 and, eventually to x² = -11 so, a number which can't be defined in the set of real numbers.
a) it's obvious at first glance that x=0 is a solution. Both the numerator and denominator are 5! = 120. b) Consider the magnitudes of the numerator and the denominator. The magnitude of the numerator is |x+2||x+3||x+4||x+5|, which the magnitude of the denominator is |x-2||x-3||x-4||x-5|. c) For any positive value of k, if x > 0, then |x+k| > |x-k|. d) For any positive value of k, if x < 0, then |x-k| > |x+k| e) In particular, from c, if x > 0, |x+2||x+3||x+4||x+5| > |x-2||x-3||x-4||x-5| f) And from d, if x < 0. |x+2||x+3||x+4||x+5| < |x-2||x-3||x-4||x-5| e) tells us that, if x >0, the magnitude of the numerator is greater than the magnitude of the denominator. Thus the underlying numbers could not possibly be equal. f) tells us the similar fact for x < 0. Thus x=0 is the only solution. This is not a problem where you want to use algebra. It's better to think about what these monomial factors mean. Ignoring the sign, the numerator is the product of the distances from the numbers -2,-3,-4, and -5, while the denominator is the product of the distances from 2, 3, 4, and 5. Clearly the former product will be smaller for x < 0, while the latter product will be smaller for x > 0.
The author got a very complicated proof, although an answer is obvious for the very first sight. Actually, the numerator at the left part of equality is always larger then the denominator. But if x=0 their moduli coincide. Then due to the number of multiplicands in the denominator are even, the products' values are coincide.
it's just a translation for russian-speaking viewers. Автор привел очень длинное доказательство, хотя ответ очевиден с самого первого взгляда. Сомножители числителя здесь всегда больше сомножителей знаменателя - сумма всегда больше разности. Но при x=0 их модули совпадают. И поскольку количество сомножителей знаменателя четное, значения произведений числителя и знаменателя совпадают. Конечно, это касается только действительного корня😂
I just saw that as a fraction which equalled 1, both parts of the fraction had to be equal. The only way that was possible was if x=0...I'm not a mathematician, and didnt understand much of the process of proving this, and dont understand why it had to be so complex when the answer was obvious at first glance. 😵💫
obviously x=0 is sol but are there others ? Lets go with brute force ... x4+14x3+71x2+154x+120 = x4-14x3+71x2-154x+120 14x3+154x=0 14x(x2+11)=0 x = 0 is only real solution x = +/-jV11 are complex solutions
10:55 Nie pisz iksa czyli x jako znaku mnożenia, tylko używaj kropki gdy mnożysz liczby. Bo X jest tu niewiadomą, więc NIE MOŻESZ TEGO X UŻYĆ JAKO ZNAKU MNOŻENIA!!!! To ewidentny błąd logiczny !!! P.S. dziękuję, że już nie raz wysłuchałeś moich rad 😊
Don't use math or algebra to solve this. Both numerator and denominator must be equal for this to work. If x = zero then top and bottom will be the same number. Multiplying 4 negative numbers on the bottom will result in a positive denominator. Whatever number the multiplications give you, divided by itself equals 1.
Multiply both sides by (x-2), and divide both sides by (x+2). ((x+3)(x+4)(x+5))/((x-3)(x-4)(x-5)) = (x-2)/(x+2) (x-2)/(x+2) | x’s cancel, leaving -2/2 = -1 Using this same logic we can conclude that (x-3)/(x+3), (x-4)/(x+4), (x-5)(x+5) all equal -1. (-1)(-1)(-1)(-1) = 1 (-1)*(-1) = 1 1 * 1 = 1 1 = 1 But we only got here by removing x from the equation. Therefore, x must equal Zero. As any other value for x will not work.
The answer is obvious by inspection, noting that the numerator factors contain +2,3,4,5 and the denominator -2,3,4,5. If you put in x=0 you get the same numerator and denominator, with the same sign as there are four negatives.
Yes. It''s so obvious, that i wonder if it's not a sort of prank made by the author that could mean: :" Heey you fools, dont step into calculations as you use to do, and like i do for solution just for the fun ". A sort of learning from the master to the beatle :"Think outside the box." Or as we could say in my native language cuz jwz born in Marseille: " Mais putaing de congg,Manu, t'as pas besoin de te prendre la tronche avé des calculs,, t'as juste à regarder que Y = -Y".
No need to all these calculations. Above multiplication should be equal to below. - times - is +, so x should be < 5. Also because above and below must be equal, solution is zero.
You should think -> how you can get 1 on the right site? if you / some number like 2000/2000 = 1 that means: (x+2) * (x+3) * (x+4) * (x+5) = (x-2) * (x-3) * (x-4) * (x-5) than think more! if you multiply 4 times negative number you will get positiv it's mean you should have x + 2 = | x - 2 | and x + 3 = | x - 3 | and so on it is mean x = 0 what is a problem to think?
Ok... but: (-4) (2) (22) = -176... that does mean sqrt of -176 divided by 4 is an imaginary number and there is no solution in real numbers... and the solution is x= (+/-) sqrt (-11 i)... !! but this procedure is great and simple...!! thanks...!!
why solving the 2nd degree equation? 2x^2 +22=0 means 2x^2=-22; x^2 as a perfect square, can never be negative, so does 2x^2 therefore .... no real solution, without solving the equation. Anyway congratulation fo r your work!
No tiene sentido hacer ninguna cuenta. Si x>0 entonces cada uno de los términos del numerador será mayor (en valor absoluto) que cada uno de los de los del denominador. Viceversa si x
Took me less than one minute to see it have to be x=0. Sometimes people overshoot the math. I have a math problem. A new bus starts every hours from New York to LA (it will not stop) and have constant speed. I drive in the opposite direction at 60 mph speed, i meet one bus every 24 minutes. How fast do the bus travel?
I am historian and finished school in 1973, 50 years ago.But I remember, that there is such thing as heuristic, very useful for math.Also I remember, that a×b=(-a)×(-b). So x=0. Having watched your explanation, I remembered an anecdote of my schooling.A teacher asks a child, how much is 2 plus 2.The child answers, it is 4.The teacters says, every fool can solve this way.Will solve it by a new easy rational way.Babby-I cant.Look 2+2= 2(1+1)= 2(2)= 2×2= 4. Do you see how it's easy.And there is more and more easier way.
Define f(x) = x*(x+1)*(x+2)*(x+3). We're looking at the quotient f(x+2)/f(x-5). Note that the product of four consecutive integers is increasing as x increases, and decreases for negative sets of integers. And since we are looking at a quotient, we can dismiss the possibility that either the numerator or denominator of this fraction is zero. So all four factors of each of the denominator and the numerator must have the same sign: and it must be different from the sign of the other. Also, the four magnitudes must be equal: i.e, {|x+2|,|x+3|,.|x+4|,|x+5|} = {|x-2|,|x-3|,|x-4|,|x-5|}. This only happens at x=0.
First of all, you must write for thèse values of x for which thé equation is définie. x must be different than 2, 3, 4 and 5. For this values, it is forbidden to use your arguments. For x different to this values, It is more simple to multiply the two members by (x-2)(x-3)(x-4)(x-5) and to expand the two members with a four degree formula. You have 14x^3+154x=0, you factorise and have 14x(x^2+11)=0. You have only x=0 as real solution, because x^2+11>0
It can be solved very strictly without calculations. 1) We fix that x ≠ 2, x ≠ 3, x ≠ 4, x ≠ 5. It's obligatory; why the author doesn't do it? 2) x = 0 is the obvious solution because the numerator and the denominator become equal if x = 0. The even number of negative multipliers in the denominator produces a positive product. 3) We have (x+2)(x+3)(x+4)(x+5) = x⁴+Bx³+Cx²+Dx+E, where B>0, C>0, D>0, E>0. 4) The denominator has the same coefficients but, sometimes, with a different sign: (x-2)(x-3)(x-4)(x-5) = x⁴-Bx³+Cx²-Dx+E. 5) The equation is x⁴+Bx³+Cx²+Dx+E = x⁴-Bx³+Cx²-Dx+E ≡ Bx³+Dx = -Bx³-Dx ≡ 2Bx³+2Dx = 0 ≡ x(x²+ (D/B)) = 0, and B>0, D>0. 6) x = 0 is a known solution, see p.2. 7) x² = -D/B has no real solutions because -D/B < 0.
X=0,(-11)^(1/2) (I am just on the beach with the calculator of iPhone without pen and paper) X=(-1*(2*3*4+2*3*5+2*4*5+3*4*5)/(2+3+4+5))^(1/2)=i*(11)^(1/2)
![ลักพาตัว เอาลูกเอิร์นกลับมา...!! [เอิร์นไดเม่]](http://i.ytimg.com/vi/f4NVKlZ5qkk/mqdefault.jpg)
In 10:07 Error! The delta = - 176 😊
Well spotted. Though this could have been avoided since
2x² + 22 = 0
2x² = -22
x² = -11
x = sqrt(-11)
😂😂
10:09 he concludes 4 times 2 times 22 = 88 ?-ok dosnt matter-
Решается в уме за 10 секунд.
A couple of innacuracies on the solution:
1. a^2-b^2 is not a perfect square, at least not in general. It is the difference of squares, and sure, there is a nice factorization formula, the one you used.
2. The product 4x2x22 is 176, not 88.
I like to estimate the good intention. But you are right to make corrections. i fould a mental solution th-cam.com/video/_biBSxDhy_0/w-d-xo.html
Знаменатель переносим вправо. Слева и справа от знака равенства полиномы имеют симметричные корни относительно 0, значит надо проверить X=0. Это корень, других корней не может быть, так как увеличение одной строны равенства приводит к уменьшению другой
The product of the distances from point X to 2, 3, 4 and 5, and
The products of the distances from point X to -2, -3, -4 and -5 are equal,
The location of the X point at this time is a problem.
Since 2, 3, 4, and 5 and -2, -3, -4, and -5 are symmetrical to each other around 0 in the coordinates,
The position of point X is 0.
Yes! This generalizes nicely.
For any positive value of x, the numerator will be a larger number than the denominator. And for any negative value of x, the numerator will be a smaller number than the denominator. Both cases make it impossible for the quotient to be equal to 1, therefore x = 0 is the only possible answer.
Excellent observation.
Try x=-1. Numerator is 24. Denominator is 360.
@@akhilranjan4460 And your point is what? Are you saying 24 is not smaller than 360? Are you saying 24/360 = 1?
2:07
@@rodjohnson2632 Sorry. I think I got a mental block at that point.
I just had an interesting solution in mind, in this ques we are solving (x-2)(x-3)(x-4)(x-5) = (x+2)(x+3)(x+4)(x+5). When we plot the graph for left hand side and right hand side, we will find 2 w shape curves in which - 5,-4,-3,-2 & 2,3,4,5 are zero. These two curves obviously will only have one intersection Point. And as we know the two curve are symmetric, so the intersection occurs in x=0.
Durch Zeichnen einer Funktion die Lösungen einer Gleichung zu finden , ist das letzte Alternative in der Mathematik.
Nicely done. I don't think you need to appeal to the discriminant though. If 2x^2 + 22 =0 it is fairly obvious that x^2 is negative, but I do see the appeal of completeness.
Bring the denominator to the other side. Expand. The fourth, second and zeroth powers of "x" fall away. Left with an equation in x and x cubed. One root is zero, left with an equation: x squared equals constant.
It is easy to check that x = 0 is a solution and it is also not difficult to see that it is the only solution.
th-cam.com/video/_biBSxDhy_0/w-d-xo.html here some simple arguments. Ithnk he waned to help beginners to have an exercise in algebra
If you won to teach math atlist you have to know maths. 4X2X22=176 and not 88
Second. For this kind of problem the first is that all this in denominator have to be different from zero. That is requirement. So please.....
Cross multiply. Distribute. The forth powers and the constant terms cancel each other. Put everything on the same side. Remains a cubic that equals 0. Factor out x. x=0 is a solution. Remains a quadratic with two complex solutions.
Mad man expresses in mad ways😂
Determinant ..cest -176 et non -88....meme si le resultat est le meme
Small mistake at the end, delta = -176. thanks for your videos.
Thanks @przejan
If you remove the forbidden values (2,3,4,5), you randomly check with 0 and 1 and you have it without developing
Super facil da 1
So easy
As the numerator and denominator have four factors, each of which is similar differing only in the sign (the factor in the numerator is a sum but in the denominator it is a difference) it easily seen that when x=0 they are the same in number but different in sign (plus in the numerator and minus in the denominator). Their product will be the same both numerical value as well as sign. The ratio is 1.
The equation should have 4 roots; as one of them is 0, you only have to solve a 3° degree equation.
Элементарно! Домножаем на знаменатель и раскрываем скобки, получается выражение вида ax^3+bx=0, у которого есть единственное решение x=0
Ну оно не единственное, о комплексных корнях мы не забываем, однако для этой задачи это верно, так как в этом видео в условии сказано, что Х принадлежит только множеству действительных чисел
Ноль как один из корней очевиден вообще без всякого решения
Автор-зануда, расписал фигню на 12 минут.Мне, историку по образованию, окончившему советскую школу ровно полтос лет тому назад очевидно, что неизвестное нуль.Вспоминается, когда занимался матекой с племянником, быстро говоришь, если равное разделить на равное, получишь...бедный ребёнок успевает выпалить (получишь) равное.Я менторским тоном : нет, получишь единицу.
Нафиг выражения "вида" . Была бы ещё пара множителей (х+1) и (х-1), и результат=-1.
10 минут для студента. Мы в 6 классе за три минуты щёлкали подобное
So simple to simply infer equality of numerator and denominator, of course 0 is one solution because applying it just leaves a positive number each side, but then for any other X it is an impossibility...
一般化して解くとこんな感じかな。
まず、a>0,x≠aとして|x+a|と|x-a|の大小関係を得る。|x+a|≧0,|x-a|≧0,y=x^2はx≧0において単調増加なので、2乗して比較しても同様の結果を得られる。
|x+a|^2 = x^2+2ax+a^2
|x-a|^2 = x^2-2ax+a^2
つまり、問題はaxと-axの大小比較に置き換えられる。
x|x-a|
が得られる。
これを変形すればa>0で
x1
となる。
ーーーーーーーーーーーーーーー
問題の式について
x1なので、式の絶対値は1より大きく不適。
x=2,3,4,5のときこの式は定義されない。
したがって、x=0
Great 🎉!!!!
Thanks!!🎉
He did it correctly, but made it unnecessarily too lengthy - start getting headache in the middle of this superfluous exercise
سهل جداً فقط اختصر كل الاقواس المتشابه والمختلفة بل الاشارة فيصبح الناتج واحد انه سهل حقا ولايحتاج هذا التعقيد
At this point 4:09 you can make the solution much faster and easier by set a= x2+7x+11 and b=x2-7x+11 instead. Then we have (a-1)(a+1)=(b-1)(b+1), means a2-1=b2-1 >> a2=b2 >> a=b or a=-b.
也只是相同結果，
I think that better way is to use a=x²+7x+11 and b=x²+7x+11. (a-1)(a+1)/(b-1)(b+1)=1; a²-1=b²-1; a²=b²...
Clever solution!
Then, X = +/- 0 two identical roots.
And X = +/- i•11^(1/2) the other two roots.
Tres bien
Красиво. Мне понравилось, как поиграли. Конечно, можно было бы покороче, но так тоже вкусно.
x = 2,3,4,5の場合、分母は0、分子は0にならないのでxの解とならない。
x != 2,3,4,5の場合
両辺に(x-2)(x-3)(x-4)(x-5)を掛けると、
(x+2)(x+3)(x+4)(x+5)=(x-2)(x-3)(x-4)(x-5)
x^nの各係数を見ると
nが偶数の場合は左辺と右辺の係数の値は同じになるため考える必要はない。
nが奇数の場合は左辺と右辺の係数の絶対値は同じで符号は逆になる。
n=3 左辺のx^3の係数は2+3+4+5＝14
n=1 左辺のxの係数は2・3・4+2・3・5+2・4・5+3・4・5＝24+30+40+60＝154
したがって方程式は14x^3+154x＝-14x^3-154x
整理すると28x^3+308x＝28x(x^2+11)=0
x=0,±sqrt(11)i
今回は実数なのでx=0
It might be useful to note that, at x=0, both the numerator and denominator equal 5!.
5! ??
5! = 5x4x3x2x1 = 120 😂😂😂
Yes true
@@widyamoe2295don’t be idiot.
He meant when x = 0.
Then num / denominator = 1
Here is my solution (of course way faster):
1) We work on the numerator. If we introduce y=x+7/2 our expression becomes: (y-3/2)(y-1/2)(y+1/2)(y+3/2)
Of course (y-1/2)(y+1/2)=y²-1/4 and (y-3/2)(y+3/2)=y²-9/4.
Now let's transform (y²-1/4)(y²-9/4). With the same idea we introduce z=y²-5/4.
Our expression becomes: (z-2)(z+2)=z²-4.
So the numerator can be written as (y²-5/4)²-4 or as [(x+7/2)²-5/4]²-4
2) The same work on the numerator would lead to the expression [(x-7/2)²-5/4]²-4
3) Our problem is equivalent to solve (for x not equal to 2, 3, 4 or 5): [(x+7/2)²-5/4]²-4=[(x-7/2)²-5/4]²-4 or [(x+7/2)²-5/4]²=[(x-7/2)²-5/4]²
We have two cases:
a) (x+7/2)²-5/4=(x-7/2)²-5/4 or (x+7/2)²=(x-7/2)²
Since x+7/2 can't be equal to x-7/2 this implies that x+7/2=-(x-7/2)=-x+7/2 then x=-x then x=0.
b) (x+7/2)²-5/4=-[(x-7/2)²-5/4] or (x+7/2)²+(x-7/2)²=5/2.
We can see that among x+7/2 and x-7/2, at least one is outside [-5/2,5/2] then its square is >25/4>5/2 and same for the sum. So we know that we don't have any real solutions.
We can still find the complex solutions:
2x²+49/2=5/2 => x²=-11 then x=sqrt(11).i or -sqrrt(11).i
Very nice 👍👍
Thank you 👍👍
Sorry, but at the determinant, in what universe does - (4 x 2 x 22) =- 88? - 4 x 2 = - 8. - 8 x 22 = - 176. And, I really don't get why that's needed anyway as the substitution result of 2x² + 22 = 0 leads to the much easier 2(x² + 11) = 0 and, eventually to x² = -11 so, a number which can't be defined in the set of real numbers.
A lot work for an answer that was immediately obvious.
Die Mathematik Aufgabe muss gelöst werden ... das bloße Vermuten ist nicht akzeptabel.
Why is this interesting for commun People?
a) it's obvious at first glance that x=0 is a solution. Both the numerator and denominator are 5! = 120.
b) Consider the magnitudes of the numerator and the denominator. The magnitude of the numerator is
|x+2||x+3||x+4||x+5|, which the magnitude of the denominator is |x-2||x-3||x-4||x-5|.
c) For any positive value of k, if x > 0, then |x+k| > |x-k|.
d) For any positive value of k, if x < 0, then |x-k| > |x+k|
e) In particular, from c, if x > 0, |x+2||x+3||x+4||x+5| > |x-2||x-3||x-4||x-5|
f) And from d, if x < 0. |x+2||x+3||x+4||x+5| < |x-2||x-3||x-4||x-5|
e) tells us that, if x >0, the magnitude of the numerator is greater than the magnitude of the denominator. Thus the underlying numbers could not possibly be equal.
f) tells us the similar fact for x < 0.
Thus x=0 is the only solution.
This is not a problem where you want to use algebra. It's better to think about what these monomial factors mean. Ignoring the sign, the numerator is the product of the distances from the numbers -2,-3,-4, and -5, while the denominator is the product of the distances from 2, 3, 4, and 5. Clearly the former product will be smaller for x < 0, while the latter product will be smaller for x > 0.
The author got a very complicated proof, although an answer is obvious for the very first sight. Actually, the numerator at the left part of equality is always larger then the denominator. But if x=0 their moduli coincide. Then due to the number of multiplicands in the denominator are even, the products' values are coincide.
it's just a translation for russian-speaking viewers. Автор привел очень длинное доказательство, хотя ответ очевиден с самого первого взгляда. Сомножители числителя здесь всегда больше сомножителей знаменателя - сумма всегда больше разности. Но при x=0 их модули совпадают. И поскольку количество сомножителей знаменателя четное, значения произведений числителя и знаменателя совпадают. Конечно, это касается только действительного корня😂
Why u r not using second bracket?
I just saw that as a fraction which equalled 1, both parts of the fraction had to be equal. The only way that was possible was if x=0...I'm not a mathematician, and didnt understand much of the process of proving this, and dont understand why it had to be so complex when the answer was obvious at first glance. 😵💫
obviously x=0 is sol but are there others ?
Lets go with brute force ...
x4+14x3+71x2+154x+120 = x4-14x3+71x2-154x+120
14x3+154x=0
14x(x2+11)=0
x = 0 is only real solution
x = +/-jV11 are complex solutions
Thank´s
Your welcome!
10:55 Nie pisz iksa czyli x jako znaku mnożenia, tylko używaj kropki gdy mnożysz liczby. Bo X jest tu niewiadomą, więc NIE MOŻESZ TEGO X UŻYĆ JAKO ZNAKU MNOŻENIA!!!! To ewidentny błąd logiczny !!! P.S. dziękuję, że już nie raz wysłuchałeś moich rad 😊
Small inconsequential error near the end… delta is -176, not -88. Otherwise, good job!
Pretty problem!
Thank you very much!
Don't use math or algebra to solve this. Both numerator and denominator must be equal for this to work. If x = zero then top and bottom will be the same number. Multiplying 4 negative numbers on the bottom will result in a positive denominator. Whatever number the multiplications give you, divided by itself equals 1.
Multiply both sides by (x-2), and divide both sides by (x+2).
((x+3)(x+4)(x+5))/((x-3)(x-4)(x-5)) = (x-2)/(x+2)
(x-2)/(x+2) | x’s cancel, leaving -2/2 = -1
Using this same logic we can conclude that (x-3)/(x+3), (x-4)/(x+4), (x-5)(x+5) all equal -1.
(-1)(-1)(-1)(-1) = 1
(-1)*(-1) = 1
1 * 1 = 1
1 = 1
But we only got here by removing x from the equation. Therefore, x must equal Zero. As any other value for x will not work.
The answer is obvious by inspection, noting that the numerator factors contain +2,3,4,5 and the denominator -2,3,4,5. If you put in x=0 you get the same numerator and denominator, with the same sign as there are four negatives.
Yes. It''s so obvious, that i wonder if it's not a sort of prank made by the author that could mean: :" Heey you fools, dont step into calculations as you use to do, and like i do for solution just for the fun ". A sort of learning from the master to the beatle :"Think outside the box." Or as we could say in my native language cuz jwz born in Marseille: " Mais putaing de congg,Manu, t'as pas besoin de te prendre la tronche avé des calculs,, t'as juste à regarder que Y = -Y".
Eye radical eleven
I might be wrong but, once you realize that aˆ2 + 2a = bˆ2+2b , then a = b. Therefore 14x=0 which yields x = 0.
Eu não faria ideeeia de por onde começar a responder, achei bastante interessante o caminho escolhido
Thanks @douglasjunior5062
are you really confident about your math knowledge or skill?
Thanks
Thanks you too. Welcome!
it's easier if you do the calculations directly
(X + 2)*(X + 3)*(X + 4)*(X + 5)
----------------------------------------------- = 1
(X - 2)*(X - 3)*(X - 4)*(X - 5)
eq to
(X + 2)*(X + 3)*(X + 4)*(X + 5)= (X - 2)*(X - 3)*(X - 4)*(X - 5)
we have
(X + 2)*(X + 3)*(X + 4)*(X + 5)=120 + 154 X + 71 X^2 + 14 X^3 + X^4
and
(X - 2)*(X - 3)*(X - 4)*(X - 5) =120 - 154 X + 71 X^2 - 14 X^3 + X^4
then (X + 2)*(X + 3)*(X + 4)*(X + 5)= (X - 2)*(X - 3)*(X - 4)*(X - 5) is eq to
120 + 154 X + 71 X^2 + 14 X^3 + X^4 - [120 - 154 X + 71 X^2 - 14 X^3 + X^4 ] = 0
from where
308 X + 28 X^3= 28 X (11 + X^2)=0
Sol = 0 , (11)^(1/2) and - (11)^(1/2)
(11)^(1/2) and - (11)^(1/2) are not integers
X=0 and X=-11 are both answer to the equation.
-4*2*22=-176.....although -88 or -176 both are not real numbers and do not have much significance here
👍
No need to all these calculations. Above multiplication should be equal to below. - times - is +, so x should be < 5. Also because above and below must be equal, solution is zero.
Квадратное уравнение получится если просто "раскрыть скобки" (т.е. привести к стандартному виду). Не вижу предмета обсуждения.
Я в уме решил за 15 секунд. 0 очевиден для этого представления, как белый день.
You should think -> how you can get 1 on the right site?
if you / some number like 2000/2000 = 1
that means:
(x+2) * (x+3) * (x+4) * (x+5) = (x-2) * (x-3) * (x-4) * (x-5)
than think more!
if you multiply 4 times negative number you will get positiv
it's mean you should have x + 2 = | x - 2 | and x + 3 = | x - 3 | and so on
it is mean x = 0
what is a problem to think?
Thanks.
No worries!
I did the long math and got x=0,±√11i with the fourth answer being null.
You can see obviously immediately that 0 is a solution. Therefore ? No need to go further.
Three restrictions:{2,3,4,5}, then x€R-{2,3,4,5}
Let a = x^2 +7x +11, b =x^2-7x+11, that would significantly simplify the equation.
Это сокращает решение в 4 раза.
Ok... but: (-4) (2) (22) = -176... that does mean sqrt of -176 divided by 4 is an imaginary number and there is no solution in real numbers... and the solution is x= (+/-) sqrt (-11 i)... !! but this procedure is great and simple...!! thanks...!!
Good point!
No need of algebra, although this is a good exercice. Just logic: x=0, symetric num/denum.
x should be different from 2,3,4,5 in other way the denominator is zero so x apart in R-(2,3,4,5).
Skipping any algebra, given symmetry of problem 0 has to be a solution
This is a solution, but not the only solution. Moving away from real numbers, you'll find that x=i*sqrt(11) is also a solution.
why solving the 2nd degree equation? 2x^2 +22=0 means 2x^2=-22; x^2 as a perfect square, can never be negative, so does 2x^2 therefore .... no real solution, without solving the equation. Anyway congratulation fo r your work!
Can get the answer through division of polynomials too.
No tiene sentido hacer ninguna cuenta. Si x>0 entonces cada uno de los términos del numerador será mayor (en valor absoluto) que cada uno de los de los del denominador. Viceversa si x
У НАС есть хорошая фраза - ОЧЕВИДНО, ЧТО....х=0 через 2 секунды)))
That accent reignites my PTSD, gained from having to deal with "technical support" people from Dell and Comcast.
Took me less than one minute to see it have to be x=0. Sometimes people overshoot the math. I have a math problem. A new bus starts every hours from New York to LA (it will not stop) and have constant speed. I drive in the opposite direction at 60 mph speed, i meet one bus every 24 minutes. How fast do the bus travel?
X= 0 is correct. But the other value is wrongly arrived by you. Check with your workings.
Тоже мне бином Ньютона! И так было понятно , что решение однозначно 0! Лишь бы бумагу переводить...
I am historian and finished school in 1973, 50 years ago.But I remember, that there is such thing as heuristic, very useful for math.Also I remember, that a×b=(-a)×(-b). So x=0. Having watched your explanation, I remembered an anecdote of my schooling.A teacher asks a child, how much is 2 plus 2.The child answers, it is 4.The teacters says, every fool can solve this way.Will solve it by a new easy rational way.Babby-I cant.Look 2+2= 2(1+1)= 2(2)= 2×2= 4. Do you see how it's easy.And there is more and more easier way.
The anecdote is good!
(x+2)/(x-2)=1
x+2=-x+2
2x=0
x=0
不知道對不對，畢竟未知數等比相乘後都是等比，最終等於1就是相等，那就不用看後面了。
4x/4x=999x/999x
數字都不用看不是嗎？只要看一組x/x=1就好了。
X=0
Lo deduje por simple inspección 😎😎😎😎
Define f(x) = x*(x+1)*(x+2)*(x+3). We're looking at the quotient f(x+2)/f(x-5). Note that the product of four consecutive integers is increasing as x increases, and decreases for negative sets of integers. And since we are looking at a quotient, we can dismiss the possibility that either the numerator or denominator of this fraction is zero. So all four factors of each of the denominator and the numerator must have the same sign: and it must be different from the sign of the other. Also, the four magnitudes must be equal: i.e, {|x+2|,|x+3|,.|x+4|,|x+5|} = {|x-2|,|x-3|,|x-4|,|x-5|}. This only happens at x=0.
Ruzsi. ROŹALIA
Mit irják nem dudm a nyelvet Anyt Tudok modani szeretlek az egés csaldodat. Sok boldoĝságt kivánok
That's a long way to go to say the numerator must equal the denominator, and the only X value that will do that is 0.
i found the zero as only solution in less than 10 seks--it is soooo obvious.
Me too but the important thing is the method
If x is not limited to real numbers. ±√11i are possible answers.
X= 0 game over
Молодец.
D=0-4*2*22=-176
Very obvious, cos the only way LHS = RHS is only if x = 0
First of all, you must write for thèse values of x for which thé equation is définie. x must be different than 2, 3, 4 and 5. For this values, it is forbidden to use your arguments.
For x different to this values, It is more simple to multiply the two members by (x-2)(x-3)(x-4)(x-5) and to expand the two members with a four degree formula. You have 14x^3+154x=0, you factorise and have 14x(x^2+11)=0.
You have only x=0 as real solution, because x^2+11>0
I got 11 as answer
Descriminant - 2*88
How about to try Zero, 0...
3
This is madness!😭
It can be solved very strictly without calculations.
1) We fix that x ≠ 2, x ≠ 3, x ≠ 4, x ≠ 5. It's obligatory; why the author doesn't do it?
2) x = 0 is the obvious solution because the numerator and the denominator become equal if x = 0. The even number of negative multipliers in the denominator produces a positive product.
3) We have (x+2)(x+3)(x+4)(x+5) = x⁴+Bx³+Cx²+Dx+E, where B>0, C>0, D>0, E>0.
4) The denominator has the same coefficients but, sometimes, with a different sign: (x-2)(x-3)(x-4)(x-5) = x⁴-Bx³+Cx²-Dx+E.
5) The equation is x⁴+Bx³+Cx²+Dx+E = x⁴-Bx³+Cx²-Dx+E ≡ Bx³+Dx = -Bx³-Dx ≡ 2Bx³+2Dx = 0 ≡ x(x²+ (D/B)) = 0, and B>0, D>0.
6) x = 0 is a known solution, see p.2.
7) x² = -D/B has no real solutions because -D/B < 0.
X=0,(-11)^(1/2)
(I am just on the beach with the calculator of iPhone without pen and paper)
X=(-1*(2*3*4+2*3*5+2*4*5+3*4*5)/(2+3+4+5))^(1/2)=i*(11)^(1/2)