Alternative way of solving this: Multiply the first equation by x to get x^2 - 5x + 1 = 0 Use the abc-formula to find the values of x; they are (5 + sqrt(21))/2 and (5 - sqrt(21))/2 Check that they are the right solutions: indeed, I don't divide by zero as both roots are nonzero. This check is necessary because in my first step I multiplied with x. Note that for either value for x, 1/x is the other value. Because the original equation was symmetric over x and 1/x. So it suffices to add up the 5th powers of both values for x. Now compute ((5 + sqrt(21))/2)^5 + ((5 - sqrt(21))/2)^5 = 1/2^5 * {(5 + sqrt(21))^5 + (5 - sqrt(21))^5} = 1/32 * {(1+1)*5^5 + (1-1)*5*5^4*sqrt(21) + (1+1)*10*5^3*sqrt(21)^2 + (1-1)*10*5^2*sqrt(21)^3 + (1+1)*5*5*sqrt(21)^4 + (1-1)*sqrt(21)^5} = 1/32 * (2*5^5 + 2*10*5^3*21 + 2*5*5*21^2) = 1/16 * (3125 + 26250 + 11025) = 40400/16 = 2525
There are many ways to solve it. I prefer an iterative way. Let Tn be an order of the function Tn=x^n+x^(-n). It can be easily proved (recursive way) that T2n+1=Tn+1*Tn-T1 and T2n=Tn^2-2!. If T1=5, then T2=5^2-2=23; T3=23*5-5=110; T4=23^2-2; T5=110*23-5=2525. So T5 is the solution to the Task.
You don't need to know any of the intermediate values of x^n + 1 / x^n. After a little algebra (expansion of 5th and 3rd powers of (x + 1/x)): x^5 + 1/x^5 = (x + 1/x)^5 - 5(x + 1/x)^3 + 5(x + 1/x) = 5^5 - 5(5^3) + 5(5) = 2525
I learned Pascal’s triangle as an 11 year old. Why would anyone solve this trivial problem any other way? Thank you for stating that fewer steps are better.
I hadn't got a clue to how to do this but I just wanted to thank you for all your questions over the year and I just hope you can think of even more for next year!
I am always not creative enough for the proper way. I however figured out that x= (5+-sqrt(21))/2) or 4,791287 or 0,2087121. From there with a calculator you arrive at 2.525. Your solution is of course far more elegant!
This one took me just a few minutes, to my surprise. I have not formally studied mathematics for decades, since I work in a profession that does not require these skills. But weirdly, I think I am better at these "math Olympiad" types of questions now, doing them just for fun & entertainment, than I ever was in the "old days." Thanks, Premath, and happy holidays.
Today I discovered this amazing platform where we can enhance our accuracy while solving mathematics which will help us in our future like in any competitive plane ... thanks and lots of best wishes to you sir ....❤️
I brute-forced my way to the correct answer lol - your solution is so easy and elegant ;) Brute-force: I rewrote the question as (x^10 + 1)/x^5 = ? I used the quadratic formula on the first equation to get x = (5 +/- sqrt(21))/2 I raised this to the 5th power - this required doing some binomial expansion in the numerator. With a calculator I simplified the answer to x^5 = (2525 +/- 551sqrt(21))/2 Then I squared THAT (which was easier!) and added 1 - this simplified to x^10 + 1 = (6,375,625 +/- 1,391,275sqrt(21))/2 I substituted my answer for x^5 and x^10 + 1 into the equation (x^10 + 1)/x^5 = ? At this point I also had my scientific calculator tell me the prime factors for each number. I then noticed that the monster fraction could be rewritten as (a(a+/-b*sqrt(21))/(a+/-b*sqrt(21)) That simplifies to just a!!! I had a = 5^2 x 101 = 2,525 This is the correct answer
first multiply the 1st eq. by x, => x^2-5x+1=0. (x-5/2)^-(21/4 )=0. x-(5/2)=(sqrt(21))/2 v x-(5/2)=-(sqrt(21))/2 so x=(5+sqrt(21))/2 or x=(5-sqrt(21))/2. now grab your calculator and fill in to get x^5+(1/(x^5))=2525 in both cases
Notice that x + 1/x = b is equivalent to: 2cosh(ln(x)) = b (for b >= 2) Solving for x, you get x = e^(arcosh(b/2)) The general solution for x^k + 1/x^k is then trivially 2cosh(k*arcosh(b/2))
Hi our worldwide teacher, as ever provide clever, clear explanation. Now you have used two principles to solve this problem : decrease the biggest powers(extent) and contrariwise for the small ones, go up to meet that powers 👍 Merry Christmas dear, with warmness from russia.
This is how I did it: I multiplied both sides of the given top equation by x, which gives you x^2-5x+1=0. Then I solved that quadratic, which gives you x = 5/2 + sqrt(21)/2 and x = 5/2 - sqrt(21)/2. Then I inserted those x-values into the equation x^5 + 1/x^5 = ?, which gives you 2525 for both x-values.
Nice one! Nice solution. I mean, of course you *CAN* also take the first equation and solve for x, find that x = (5 +/- sqrt(21))/2, then take those terms and their reciprocals, raise them to the power 5 and find 2525 that way. But that's just boring, really.... 🤣🤣🤣
Pretty simple, Let x=c,1/x=d and t^2+at+b=0,(t=c,d) Applying Vieta formula, a=-5 & b=1 , Applying Newton sum if p_n=c^n+d^n p_n+a(p_n-1)+b(p_n-2)=0 p_2=5(p_1)-(p_0)=23 p_3=5(p_2)-(p_1)=110 p_4=5(p_3)-(p_2)=527 p_5=5(p_4)-(p_3)=2525 Ans :)
much easier way. so take the original equation1 X+1/X = 5 . you can square both sides and reduce to get equation2 X2+1/X^2=23. you can multiply left side of this equation2 with X+1/x and the right side with 5 (since they equal each other) to get X3+(X+1/x)+1/X^3= 23x5 and you replace the X+1/X with 5 so you get equation3 X3+1/X^3=110. now you can multiply the equation3 with equation2 to get (X3+1/X3)(X2+1/X^2)=110x23. multiply and reduce using the above method so you get X5 +(X+1/X)+1/X^5 = 2530 and you know the part in bracket is 5, so then subtract 5 from each side to get X5+1/X^5= 2525
Not sure this needed second powers or fancier expansions. Expand (x + 1/x)^5 and (x + 1/x)^3 and pair similar to de Moivre's theorem of complex powers. 3125 - 5(110) - 10(5) = 2525
@@XLatMaths I only used Pascal’s triangle and got the correct answer very quickly. What I was implying was that one concept (Pascal’s Triangle) was sufficient.
Yes! Let S(n) = x^n+1/(x^n). Then S(1) = 5 and S(0) = x^0+1/(x^0) = 2. Now if you multiply S(n+1) and S(1) (which is 5), you get S(n+1)S(1) = x^(n+2)+x(1/x^(n+1))+(1/x)(x^(n+1))+1/x^(n+2) = S(n+2) + x(1/x)S(n) = S(n+2) + S(n). Therefore S(n+2) = 5S(n+1) - S(n). Since S(0) and S(1) are both integers, all S(n) with n>1 are integers as well. You can use this formula to calculate S(5), you will get the same result 2525!
My mind is not trained to go where you go on this and many of your problems. Instead of starting with the x^5+1/x^5 as you do, my mind tells me to start with the much simpler looking equation x+1/x=5 and solve that one for x, which ends up giving solutions of x = 2.5 +- sqrt(21)/2. Then, I plug in these values to the x^5 + 1/x^5 expression to get the answer. I'm trying to figure out how your mind figures to initiate these problems in the way you do. Thanks as always.
There are many many ways to take care of this problem. In my humble opinion, manipulating x^5 + 1/x^5 to our advantage would be a more efficient and elegant way! Thank you for your feedback 😃 Happy holidays, merry christmas, and happy new year! You are awesome my friend😀
@@bentels5340 I'll try to keep that in mind next time. I want to as a mathematician, learn to do things more efficiently and your suggestion seems like a good step in that direction.
@@PreMath I agree with your opening opinion. What I'm trying to do is figure out how to train my mind to have the instinct to try your way of doing problems like this. Thanks.
alternative method using the quadratic formula: x + 1/x = 5 notice that x != 0, there are no solutions when x is zero so it's safe to multiply both sides by x xx + 1 = 5x xx - 5x + 1 = 0 apply the quadratic formula: x = (5 +- sqrt(25 - 4) )/2 x = 2.5 +- sqrt(21)/2 because we were solving an equation of the form x + 1/x = something, the two solutions by necessity will be inverses meaning that 5 + (sqrt21)/2 = 1/ (5 - (sqrt21)/2) so x + 1/x = 5/2 + sqrt21/2 + 5/2 - sqrt21/2 now use that info to calculate x^5 + 1/x^5 x = (5 + sqrt21)/2 x^5 = (5 + sqrt21)^5 / 32 1/x = (5 - sqrt21)/2 (1/x)^5 = (5 - sqrt21)^5 / 32 thus x^5 + 1/x^5 = (1/32) * ( (5 + sqrt21)^5 + (5 - sqrt21)^5 ) it's easy enough to expand the polynomials using pascal's triangle (1 5 10 10 5 1) and calculate the result
Wait there should be a general formula for this , what if (x+1/x) is given and it is asked for (x^2000+1/x^2000) or what if the exponent's aren't same . I am not able to find a general equation for this , plz tell me .
Umm, yeah, there is... Won't help much though... m^A + n^A = (m+n)^A - R Where R is the Ath degree polynomial expansion of Newton without the first and last term (which are, coincidentally, on the other side of the equals sign).
Too complicated. The LHS of the 2nd equation suggests you take both sides of the 1st equation to the 5th power using the binomial expansion. Interchanging the two sides, this is: 5^5 = (x +1/x)^5 = x^5 + 5x^3 + 10 x + 10/x + 5/x^3 + 1/x^5. So x^5 + 1/x^5 = 5^5 - 10(x+1/x) - 5(x^3 +1/x^3); 5^5 = 3125. We know that x+1/x =5, and all that remains is to find x^3 + 1/x^3, which we can get by now instead raising the 1st equation to the 3rd power: 5^3 = (x+1/x)^3 = x^3 + 3x + 3/x + 1/x^3 = x^3 + 1/x^3 + 3(5) and so x^3 + 1/x^3 = 5^3 - 3(5) = 125 - 15 = 110 Then x^5 + 1/x^5 = 3125 - 10(5) - 5(110) = 3125 - 50 -550 = 2525.
I'm learning lots of thinG and MATH is my Favorite subject all time. Your lecture style is nicE and every point easly to understand. Keep it up SIR !
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thinG
nicE
I agree !!
Alternative way of solving this:
Multiply the first equation by x to get x^2 - 5x + 1 = 0
Use the abc-formula to find the values of x; they are (5 + sqrt(21))/2 and (5 - sqrt(21))/2
Check that they are the right solutions: indeed, I don't divide by zero as both roots are nonzero. This check is necessary because in my first step I multiplied with x.
Note that for either value for x, 1/x is the other value. Because the original equation was symmetric over x and 1/x. So it suffices to add up the 5th powers of both values for x.
Now compute ((5 + sqrt(21))/2)^5 + ((5 - sqrt(21))/2)^5
= 1/2^5 * {(5 + sqrt(21))^5 + (5 - sqrt(21))^5}
= 1/32 * {(1+1)*5^5 + (1-1)*5*5^4*sqrt(21) + (1+1)*10*5^3*sqrt(21)^2 + (1-1)*10*5^2*sqrt(21)^3 + (1+1)*5*5*sqrt(21)^4 + (1-1)*sqrt(21)^5}
= 1/32 * (2*5^5 + 2*10*5^3*21 + 2*5*5*21^2)
= 1/16 * (3125 + 26250 + 11025)
= 40400/16 = 2525
Yep, I did it just like you, it worked.
i did it that way too...
There are many ways to solve it. I prefer an iterative way. Let Tn be an order of the function Tn=x^n+x^(-n). It can be easily proved (recursive way) that T2n+1=Tn+1*Tn-T1 and T2n=Tn^2-2!. If T1=5, then T2=5^2-2=23; T3=23*5-5=110; T4=23^2-2; T5=110*23-5=2525. So T5 is the solution to the Task.
You don't need to know any of the intermediate values of x^n + 1 / x^n.
After a little algebra (expansion of 5th and 3rd powers of (x + 1/x)):
x^5 + 1/x^5 = (x + 1/x)^5 - 5(x + 1/x)^3 + 5(x + 1/x)
= 5^5 - 5(5^3) + 5(5) = 2525
I learned Pascal’s triangle as an 11 year old. Why would anyone solve this trivial problem any other way?
Thank you for stating that fewer steps are better.
i went to calculate the value of X with discriminant and then compute it at power 5 but your solution is far more elegant
I hadn't got a clue to how to do this but I just wanted to thank you for all your questions over the year and I just hope you can think of even more for next year!
Happy new year!
Thank you for your feedback! Cheers!
You are awesome David😀
Marvelous! Very grateful and appreciated.
Good job PreMath !! You made learning math easy !!
I love your method. I’d have used the quadratic formula myself to solve for X and gone from there.
I am always not creative enough for the proper way. I however figured out that x= (5+-sqrt(21))/2) or 4,791287 or 0,2087121. From there with a calculator you arrive at 2.525. Your solution is of course far more elegant!
So nice of you Philip
Thank you for your feedback 😃
You are awesome😀
first thing we did with the binomials was solve for x in x^2+px+q=0, less elegant but also fewer steps to make errors.
Same I did it this way
That's great !! I agree !!
Thankyou for given information.
Thanks sir.
Simple but more interesting.
Muchas gracias amigo, excelente la explicación. Saludos.
Questão difícil, porém com a sua explicação ficou fácil entender. Obrigado 🙏 mestre.
This one took me just a few minutes, to my surprise. I have not formally studied mathematics for decades, since I work in a profession that does not require these skills. But weirdly, I think I am better at these "math Olympiad" types of questions now, doing them just for fun & entertainment, than I ever was in the "old days." Thanks, Premath, and happy holidays.
Thank you! 😃
Happy holidays, merry christmas, and happy new year!
You are awesome JR😀
very well done, thanks for sharing, have a merry Christmas
Thanks, you too!
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You are awesome my dear friend😀
Very nice sir ji.
Thanks 😊
Most welcome Snehal
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You are awesome 😀
Today I discovered this amazing platform where we can enhance our accuracy while solving mathematics which will help us in our future like in any competitive plane ... thanks and lots of best wishes to you sir ....❤️
Welcome aboard Avann
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Love and prayers from the USA!
very well explained, thanks for sharing, have a merry Christmas
Thank you! 😃
Happy holidays, merry christmas, and happy new year!
You are awesome 😀
Love and prayers from the USA!
Hey my friend! nice problem to share haha Enjoyed the video! Keep rocking!
Thanks for the visit
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You are awesome my dear friend😀
Thank You for posting 😃
So nice of you Sana
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I brute-forced my way to the correct answer lol - your solution is so easy and elegant ;)
Brute-force:
I rewrote the question as (x^10 + 1)/x^5 = ?
I used the quadratic formula on the first equation to get x = (5 +/- sqrt(21))/2
I raised this to the 5th power - this required doing some binomial expansion in the numerator. With a calculator I simplified the answer to x^5 = (2525 +/- 551sqrt(21))/2
Then I squared THAT (which was easier!) and added 1 - this simplified to x^10 + 1 = (6,375,625 +/- 1,391,275sqrt(21))/2
I substituted my answer for x^5 and x^10 + 1 into the equation (x^10 + 1)/x^5 = ?
At this point I also had my scientific calculator tell me the prime factors for each number. I then noticed that the monster fraction could be rewritten as (a(a+/-b*sqrt(21))/(a+/-b*sqrt(21))
That simplifies to just a!!! I had a = 5^2 x 101 = 2,525
This is the correct answer
nice
nice question sir thank u
So nice of you Nico
Thank you for your feedback 😃
You are awesome😀
first multiply the 1st eq. by x, => x^2-5x+1=0. (x-5/2)^-(21/4 )=0. x-(5/2)=(sqrt(21))/2 v x-(5/2)=-(sqrt(21))/2 so x=(5+sqrt(21))/2 or x=(5-sqrt(21))/2. now grab your calculator and fill in to get x^5+(1/(x^5))=2525 in both cases
Pen and paper and wisdom are more elegant weapons than a calculator.
Notice that x + 1/x = b is equivalent to:
2cosh(ln(x)) = b (for b >= 2)
Solving for x, you get x = e^(arcosh(b/2))
The general solution for x^k + 1/x^k is then trivially 2cosh(k*arcosh(b/2))
Hi our worldwide teacher, as ever provide clever, clear explanation.
Now you have used two principles to solve this problem : decrease the biggest powers(extent) and contrariwise for the small ones, go up to meet that powers 👍
Merry Christmas dear, with warmness from russia.
Thank you! 😃
Happy holidays, merry christmas, and happy new year!
You are awesome 😀
Love and prayers from the USA!
Literally I try to solve this question for 23 min and in 24th min I solve it
...... I love your videos.....
Yea did it in my head
Nice
Excellent
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You are awesome Susen😀
yes, first one is quadratic, I can solve it since high school. Then you just substitute in the second equation :)
This is how I did it: I multiplied both sides of the given top equation by x, which gives you x^2-5x+1=0. Then I solved that quadratic, which gives you x = 5/2 + sqrt(21)/2 and x = 5/2 - sqrt(21)/2. Then I inserted those x-values into the equation x^5 + 1/x^5 = ?, which gives you 2525 for both x-values.
Nice one! Nice solution.
I mean, of course you *CAN* also take the first equation and solve for x, find that x = (5 +/- sqrt(21))/2, then take those terms and their reciprocals, raise them to the power 5 and find 2525 that way.
But that's just boring, really.... 🤣🤣🤣
I'm glad somebody else thought this way. That's how I did it.
Solving for x and then raising to the power 5 is a lot more work!
Wow i always feel your problems is too much easy but this time, it actually hard xD, keep it up!
Thank you for your feedback! Cheers!
You are awesome Van😀
This is very important for my exam 🙏🏿🙏🏿🙏🏿🙏🏿 Thank you sir
All the best!
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You are awesome 😀
Use Binomial expansion on (a+b)^5 and (a+b)^3. It'll save a little time but the idea is the same.
Великолепно!
Pretty simple,
Let x=c,1/x=d
and t^2+at+b=0,(t=c,d)
Applying Vieta formula, a=-5 & b=1
, Applying Newton sum if p_n=c^n+d^n
p_n+a(p_n-1)+b(p_n-2)=0
p_2=5(p_1)-(p_0)=23
p_3=5(p_2)-(p_1)=110
p_4=5(p_3)-(p_2)=527
p_5=5(p_4)-(p_3)=2525 Ans :)
Cool
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nice!
That's our Premath 😊👍
So nice of you Sameer
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You are awesome as always😀
clear denominator;
move everything to the left.
x^2-5x+1=0 use quad formula
then use a calculator
much easier way. so take the original equation1 X+1/X = 5 . you can square both sides and reduce to get equation2 X2+1/X^2=23. you can multiply left side of this equation2 with X+1/x and the right side with 5 (since they equal each other) to get X3+(X+1/x)+1/X^3= 23x5 and you replace the X+1/X with 5 so you get equation3 X3+1/X^3=110. now you can multiply the equation3 with equation2 to get (X3+1/X3)(X2+1/X^2)=110x23. multiply and reduce using the above method so you get X5 +(X+1/X)+1/X^5 = 2530 and you know the part in bracket is 5, so then subtract 5 from each side to get X5+1/X^5= 2525
Very good explanation i did (a+b)^2^2x(a+b)
Great job!
Excellent Mahalakshmi
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So, I took first equation and multiple both sides by x turning it into quadratic, solved for x,
and then substituted into the second equation.
My favourite subject math and geometry
Also, x^3+1/x^3=(x+1/x)(x^2-1+1/x^2)=5(23-1)=5×22=110
Multiply (x-square + 1 / x-square) by (x-cube +1 / x-cube) to simplify the last part.
32.31
Not sure this needed second powers or fancier expansions.
Expand (x + 1/x)^5 and (x + 1/x)^3 and pair similar to de Moivre's theorem of complex powers. 3125 - 5(110) - 10(5) = 2525
Learn Pascal’s Triangle
@@dianeweiss4562 I have. My answer is correct.
@@XLatMaths I only used Pascal’s triangle and got the correct answer very quickly.
What I was implying was that one concept (Pascal’s Triangle) was sufficient.
Nice example. But can you prove that for any N>0 the result x^N+(1/x)^N is integer?
Yes! Let S(n) = x^n+1/(x^n). Then S(1) = 5 and S(0) = x^0+1/(x^0) = 2. Now if you multiply S(n+1) and S(1) (which is 5), you get S(n+1)S(1) = x^(n+2)+x(1/x^(n+1))+(1/x)(x^(n+1))+1/x^(n+2) = S(n+2) + x(1/x)S(n) = S(n+2) + S(n). Therefore S(n+2) = 5S(n+1) - S(n). Since S(0) and S(1) are both integers, all S(n) with n>1 are integers as well.
You can use this formula to calculate S(5), you will get the same result 2525!
Let f(n) = x^n + (1/x) ^n.
By checking f(n+1) = f(n) f(1) - f(n-2), mathematical induction can do the rest😜
2+1/2=2.5 its not an integer N=1>0
My mind is not trained to go where you go on this and many of your problems. Instead of starting with the x^5+1/x^5 as you do, my mind tells me to start with the much simpler looking equation x+1/x=5 and solve that one for x, which ends up giving solutions of x = 2.5 +- sqrt(21)/2. Then, I plug in these values to the x^5 + 1/x^5 expression to get the answer. I'm trying to figure out how your mind figures to initiate these problems in the way you do. Thanks as always.
"For exponents greater than 2, always factor to get them smaller".
Doesn't always work, but it's a start.
There are many many ways to take care of this problem. In my humble opinion, manipulating x^5 + 1/x^5 to our advantage would be a more efficient and elegant way!
Thank you for your feedback 😃
Happy holidays, merry christmas, and happy new year!
You are awesome my friend😀
@@bentels5340 I'll try to keep that in mind next time. I want to as a mathematician, learn to do things more efficiently and your suggestion seems like a good step in that direction.
@@bentels5340 You make a good suggestion. I'll try to keep that in mind for next time. Thanks.
@@PreMath I agree with your opening opinion. What I'm trying to do is figure out how to train my mind to have the instinct to try your way of doing problems like this. Thanks.
Set play back speed x2, to save time
My answer is 2525
Bcz X came out to be {(5/2) +,- (√21/2)}
And for both" +,-" the answer was same.
alternative method using the quadratic formula:
x + 1/x = 5
notice that x != 0, there are no solutions when x is zero
so it's safe to multiply both sides by x
xx + 1 = 5x
xx - 5x + 1 = 0
apply the quadratic formula:
x = (5 +- sqrt(25 - 4) )/2
x = 2.5 +- sqrt(21)/2
because we were solving an equation of the form x + 1/x = something, the two solutions by necessity will be inverses
meaning that 5 + (sqrt21)/2 = 1/ (5 - (sqrt21)/2)
so x + 1/x = 5/2 + sqrt21/2 + 5/2 - sqrt21/2
now use that info to calculate x^5 + 1/x^5
x = (5 + sqrt21)/2
x^5 = (5 + sqrt21)^5 / 32
1/x = (5 - sqrt21)/2
(1/x)^5 = (5 - sqrt21)^5 / 32
thus x^5 + 1/x^5 = (1/32) * ( (5 + sqrt21)^5 + (5 - sqrt21)^5 )
it's easy enough to expand the polynomials using pascal's triangle (1 5 10 10 5 1) and calculate the result
Very interesting brother. Good luck!!!!!!!!!!!!
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Thank you for your continued love and support. Cheers!
You are awesome AKD😀
#binomial #polynomial
Sir my answer is 2525 .
I have done in the same way.it was easy.
Excellent Badal
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You are awesome😀
Could the answer not be 2508 since the first equation also has a solution of .209?
You could have just used the first equation to solve for x first. Then substitute x into the second equation.
Good afternoon, Sir Teacher! Here's my way of solution.
Let's introduce sum of fifth powers by the multiplying sums of two following powers. We see that we deal with 5, so it would be better to split it into 2 and 3, because there may hide a value that we know:
x⁵ + 1/x⁵ = (x² + 1/x²) • (x³ + 1/x³) - (x + 1/x) = x⁵ + x²/x³ + x³/x² + 1/x⁵ - x - 1/x = x⁵ + (1/x + x) + 1/x⁵ - (x + 1/x) = (x² + 1/x²)(x³ + 1/x³) - 5;
Let's introduce 2nd and 3rd powers by the same way through known value:
(x² + 1/x²) = x² + 2 + 1/x² - 2 = (x + 1/x)² - 2 = 5² - 2 = 23;
(x³ + 1/x³) = (x + 1/x)(x² + 1/x² - 1) = 5 • (23 - 1) = 5 • 22 = 110;
By the way:
x⁵ + 1/x⁵ = (x² + 1/x²)(x³ + 1/x³) - 5 = 23 • 110 - 5 = 23 • (10 + 1) • 10 - 5 = (230 + 23) • 10 - 5 = 253 • 10 - 5 = 2530 - 5 = 2525;
That's our final answer ;)
Answer: x⁵ + 1/x⁵ = 2525;
x^3+1/x^3=125-15=110.x^21/x^2=25-10=15.now : x^5+1/x^5=110×15-5=1645.ans
So we know a formula:
a^5 + b^5 = (a + b)(a^4 - ab^3 + a^2*b^2 - a^3*b + b^4)
Let a = x and b = 1/x:
x^5 + 1/x^5 = (x + 1/x)(x^4 - x/x^3 + x^2/x^2 - x^3/x + 1/x^4)
x^5 + 1/x^5 = 5(x^4 + 1/x^4 - (x^2 + 1/x^2) + 1)
Nice. But what about these quartic and quadratic terms? Can we handle them? Yes we can!
First, the quadratics:
(x + 1/x)^2 = x^2 + 1/x^2 + 2x/x = x^2 + 1/x^2 + 2
5^2 = x^2 + 1/x^2 + 2
25 = x^2 + 1/x^2 + 2
x^2 + 1/x^2 = 23 [E]
And the quartics. Square [E]:
x^4 + 1/x^4 + 2 = 529
x^4 + 1/x^4 = 527
And here we go.
x^5 + 1/x^5 = 5(527 - 23 + 1) = 5(505) = 2525
(That's one of the most beautiful answers I've seen in a while)
Nice question thanks a lot
Welcome Pranav
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1st Solution
Bravo
can it be x.x -5x +1=0 and find x?
Let me say something without watching the video.
To me it looks simple. Take the 5th power on both sides and then use binomial expression.
It's rare moment to read smart comments on youtube
My step2 is to multiple (x+1/x)*(x^2+1/x^2) 🙂
9.4
6th comment and ans : 2525
Excellent
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You are awesome Jayant😀
Wait there should be a general formula for this , what if (x+1/x) is given and it is asked for (x^2000+1/x^2000) or what if the exponent's aren't same .
I am not able to find a general equation for this , plz tell me .
Umm, yeah, there is... Won't help much though...
m^A + n^A = (m+n)^A - R
Where R is the Ath degree polynomial expansion of Newton without the first and last term (which are, coincidentally, on the other side of the equals sign).
@@bentels5340 Thank you
976//4.01.22.
Simple x^5+1/x^5=2525.
I got advantage coz i know binomial expansion theorem , pretty easy for an asian
It’s quadartic
Hi...
Hello Silambu
You are awesome 😀
@@PreMath how are you friend..
It's value is 2525
#distributiveproperty
x + (1 ÷ x) = 5
X+ 1/x=5
X² +1=5x
X² -5x +1=0
Solve x and then put it in the formula
You can also solve it that way
30 sec I solve it😂😂. Ssc cgl exam questions
Excellent
Thank you for your feedback 😃
You are awesome😀
a = x
b = 1 ÷ x
Just answer the X²-5X+1=0 equality and you have the X value....
23
Okey but what about :
x⁵+1/x⁵=x⁵+(1/X)⁵={X+(1/X)}⁵=(X+1/X )5=5⁵
2525. Grande!!!! Buon natale
Super Giuseppe. Buon Natale!
Grazie per il vostro continuo amore e supporto. Saluti!
Sei fantastico👍
=2525
Still don't understand this math. 😕
2525
Haha I shld have tried manipulated a bit. I gave up too easily
No worries
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You are awesome Garrick😀
Ahhhhhhhh
Some jee advanced problems too xD
2525 by an easy method.
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You are awesome Ranveer😀
I have solved it in 46 seconds. It was a very easy one. However, the answer is 2525.
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You are awesome Mustafiz😀
In 2minutes I solved sir answer is 2525🤓🤓
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You are awesome Abhishek😀
7777777
Too complicated.
The LHS of the 2nd equation suggests you take both sides of the 1st equation to the 5th power using the binomial expansion.
Interchanging the two sides, this is:
5^5 = (x +1/x)^5 = x^5 + 5x^3 + 10 x + 10/x + 5/x^3 + 1/x^5.
So
x^5 + 1/x^5 = 5^5 - 10(x+1/x) - 5(x^3 +1/x^3); 5^5 = 3125.
We know that x+1/x =5, and all that remains is to find x^3 + 1/x^3, which we can get by now instead raising the 1st equation to the 3rd power:
5^3 = (x+1/x)^3 = x^3 + 3x + 3/x + 1/x^3 = x^3 + 1/x^3 + 3(5)
and so
x^3 + 1/x^3 = 5^3 - 3(5) = 125 - 15 = 110
Then
x^5 + 1/x^5 = 3125 - 10(5) - 5(110) = 3125 - 50 -550 = 2525.
2525