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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.
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.
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
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 ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-_biBSxDhy_0.html
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.
ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-_biBSxDhy_0.html here some simple arguments. Ithnk he waned to help beginners to have an exercise in algebra
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.
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.
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.
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.
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.
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.
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 😊
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😂
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
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. 😵💫
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...!!
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".
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.....
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?
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.
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.
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
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!
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?
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.
صحيح انا عرفت الحل دير المساواة بين البسط و المقام وتخرج العامل المشترك لي هو x وتحسب الموازانة يخرجلك البسط هو نفسو المقام وبالتالي فقيمة الاكس منعدمة
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.
