That's a good and simple method. Here is the long method that people likes to tend to do algebraically, which is to multiply both sides by the LCD, which is (x-2)(x-3)(x-4)(x-5) since this would take up the time consumption if it were to be on the exam. If you do this, we get (x-10)(x-3)(x-4)(x-5)+(x-9)(x-2)(x-4)(x-5)+(x-8)(x-2)(x-3)(x-5)+(x-7)(x-2)(x-3)(x-4)=-4(x-2)(x-3)(x-4)(x-5). However, if you multiply it out term by term and using simplification and doing things algebraically by multiplying out the polynomials going from there, we get 4x^4-76x^3+504x^2-1396x+1368=-4x^4+56x^3-284x^2+616x-480. Setting that equal to 0 gives us 8x^4-132x^3+788x^2-2012x+1848=0. Divide everything by 4 is 2x^4-33x^3+197x^2-503x+462=0. You can use the rational roots theorem by finding the factors of 462 and 2 and then divide each of the factors. When you do that, we get that x=6 is one of the solutions. After you find another factor, it shows that x=7/2 is also a solution. A factored form comes out to be 4(x-6)(2x-7)(x^2-7x+11)=0. However, there are two more solutions since x^2-7x+11=0 can be used using the quadratic formula, which is x=(7±√(5))/2. Yes, I know there is a 4 in the front, but this is just before I divide everything by 4 and using the whole complete factored form.
I started by long-dividing each fraction to get: [1 - 8/(x-2)] + [1 - 6/(x-3)] + [1 - 4/(x-4)] + [1 - 2/(x-5)] = -4 . Then gather all constants to the right. Unfortunately, it gets *real* messy from that point on... so I sub'd u=x-3 to make getting common denominators and multiplying out the numerators, plus gathering all like terms a good bit nicer, resulting in the quartic: 2u^4 - 9u^3 + 8u^2 + 4u -3 = 0 . Luckily, using the rational root theorem and synthetic division I get 2 roots of u = 1/2 and u = 3. The resulting quadratic u^2 - u -1 gives us the last 2 roots (using the quadratic formula). Back-subbing gets the same 4 answers as in the video. Like 4 times as much work. :-/ But I did get the answers. LOL! :-) I absolutely *love* watching these videos for all the interesting ways of solving problems. I learn **a lot**. Thx!
Выглядит как уравнение 4-ой степени, так что решить методом Феррари можно. Но я бы сделал замену t = (8x - (2+3+4+5+7+8+9+10))/8 = (8x - 48)/8 = x - 6 = t. В таком случае уравнение станет более красивым. Ну а потом, чисто из любопытства подставлю x = 6 и получается тождество. Один корень нашли и на него можно поделить уравнение 4-ой степени, чтобы получить уравнение 3-ей степени. Ну а дальше методом Кардано)
l think X can only equal to 6. l like the way 1 was added to each terms, quite clever. Videos are greatly amazing! x - 10/x - 2(x - 2) + x - 9/x - 3(x - 3) + x - 8/x - 4(x - 4) + x - 7/x - 5(x - 5) = --4 -- multiply each terms by the denominator as shown x - 10 + x - 9 + x - 8 + x - 7 = --4( x - 2 + x - 3 + x - 4 + x - 5)/4 - Result from above, we are taking the average of the given factors( x - 2, x - 3, x - 4 and x - 5 ) 4x - 34 = -- 4( 4x - 14/4 ) --- result when calculated the above 4x - 34 = 14 - 4x 4x - 34 - (14 - 4x ) = 0 4x - 34 - 14 + 4x = 0 8x - 48 = 0 ; x = 48/8 = 6, x = 6
here's how i did it, without looking. to get a negative number, then at least some of the fractions must be negative. and to be negative, we note that the top is less than the bottom of each fraction. sp we must have a number that makes the top negative and the bottom positive. now for integers, on the right most fraction, x=6 makes -1/1 and in fact all the other fractions are -1 at x=6, so that is certainly a solution. try x=7, … -3/5,-2/4,-1/3,0/2 sums to not -4 at which point i give up and assume x=6 is the only solution. sorry.. Lazy tonight.
Thanks for video. Good luck!!!!!
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@@PreMath qqqqqqqqqqqq
@@PreMath q
The key to this question is to assign "1" to each of 4 rational terms, and the right side results in "0".
Much thanks for your best efforts.
Very good!
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That's a good and simple method. Here is the long method that people likes to tend to do algebraically, which is to multiply both sides by the LCD, which is (x-2)(x-3)(x-4)(x-5) since this would take up the time consumption if it were to be on the exam. If you do this, we get (x-10)(x-3)(x-4)(x-5)+(x-9)(x-2)(x-4)(x-5)+(x-8)(x-2)(x-3)(x-5)+(x-7)(x-2)(x-3)(x-4)=-4(x-2)(x-3)(x-4)(x-5). However, if you multiply it out term by term and using simplification and doing things algebraically by multiplying out the polynomials going from there, we get 4x^4-76x^3+504x^2-1396x+1368=-4x^4+56x^3-284x^2+616x-480. Setting that equal to 0 gives us 8x^4-132x^3+788x^2-2012x+1848=0. Divide everything by 4 is 2x^4-33x^3+197x^2-503x+462=0. You can use the rational roots theorem by finding the factors of 462 and 2 and then divide each of the factors. When you do that, we get that x=6 is one of the solutions. After you find another factor, it shows that x=7/2 is also a solution. A factored form comes out to be 4(x-6)(2x-7)(x^2-7x+11)=0. However, there are two more solutions since x^2-7x+11=0 can be used using the quadratic formula, which is x=(7±√(5))/2. Yes, I know there is a 4 in the front, but this is just before I divide everything by 4 and using the whole complete factored form.
Thanks Justin for sharing! Cheers!
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Your channel is best for practicing maths. Very very very................ Nice!!!!
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Very interesting question👍
Thanks for sharing😊
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I started by long-dividing each fraction to get: [1 - 8/(x-2)] + [1 - 6/(x-3)] + [1 - 4/(x-4)] + [1 - 2/(x-5)] = -4 . Then gather all constants to the right. Unfortunately, it gets *real* messy from that point on... so I sub'd u=x-3 to make getting common denominators and multiplying out the numerators, plus gathering all like terms a good bit nicer, resulting in the quartic: 2u^4 - 9u^3 + 8u^2 + 4u -3 = 0 . Luckily, using the rational root theorem and synthetic division I get 2 roots of u = 1/2 and u = 3. The resulting quadratic u^2 - u -1 gives us the last 2 roots (using the quadratic formula). Back-subbing gets the same 4 answers as in the video. Like 4 times as much work. :-/ But I did get the answers. LOL! :-)
I absolutely *love* watching these videos for all the interesting ways of solving problems. I learn **a lot**. Thx!
Thanks for solving this equation in such an elegant manner.
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Professor, thanks for sharing! Good luck.
I find it interesting that two of the solutions are symmetrical about one of the others.
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Выглядит как уравнение 4-ой степени, так что решить методом Феррари можно. Но я бы сделал замену t = (8x - (2+3+4+5+7+8+9+10))/8 = (8x - 48)/8 = x - 6 = t. В таком случае уравнение станет более красивым. Ну а потом, чисто из любопытства подставлю x = 6 и получается тождество. Один корень нашли и на него можно поделить уравнение 4-ой степени, чтобы получить уравнение 3-ей степени. Ну а дальше методом Кардано)
Very helpful video, thank you very much keep up the great content!
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Very nice technic in math, thank you
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l think X can only equal to 6. l like the way 1 was added to each terms, quite clever. Videos are greatly amazing!
x - 10/x - 2(x - 2) + x - 9/x - 3(x - 3) + x - 8/x - 4(x - 4) + x - 7/x - 5(x - 5) = --4 -- multiply each terms by the denominator as shown
x - 10 + x - 9 + x - 8 + x - 7 = --4( x - 2 + x - 3 + x - 4 + x - 5)/4 - Result from above, we are taking the average of the given factors( x - 2, x - 3, x - 4 and x - 5 )
4x - 34 = -- 4( 4x - 14/4 ) --- result when calculated the above
4x - 34 = 14 - 4x
4x - 34 - (14 - 4x ) = 0
4x - 34 - 14 + 4x = 0
8x - 48 = 0 ; x = 48/8 = 6, x = 6
Your videos has increased my interest of Mathematics.
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nice solved👏👏👏
bravo
Nice problem sir , thanks for your problem!
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V nice question vnicely solved
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Brilliant! Thank you
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Thanks was a great question
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Interesting problem 👏👏👏
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You are a brilient mathamatician sir
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does this technique has any name?
Oh my God Thanks very Nice🙏🙏🙏🌹🌹🌹
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Awesome
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very well explained
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Very good!
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here's how i did it, without looking. to get a negative number, then at least some of the fractions must be negative.
and to be negative, we note that the top is less than the bottom of each fraction.
sp we must have a number that makes the top negative and the bottom positive.
now for integers, on the right most fraction, x=6 makes -1/1 and in fact all the other fractions are -1 at x=6, so that is certainly a solution.
try x=7, … -3/5,-2/4,-1/3,0/2 sums to not -4 at which point i give up and assume x=6 is the only solution.
sorry.. Lazy tonight.
Perfect
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各加 1 真的是太神了
Only add 4 both sides equation.
A good math problem
🙏🏼 answer sharing x=6
Jayant Sahay 45 minutes ago comments that X=6. I plug in the number 6 for X and wallah , it works...equation equals -4 . What's goin on here?
Please disregard this question...I see why there are multiple solutions. 🙂
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I will back not Immediately but definately
Ans . x = 6
Nice question but i managed to find the one solution 6 but incomplete answer is wrong answer
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Outro show!
Solution by insight
-1-1-1-1=-4
x=6
my dumbass would just take the LCM of everything 😔
But the answer is 77/15.
x=6
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Premath!
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~10 seconds
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