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Spencer's Academy
Приєднався 6 бер 2024
This channel is devoted to equipping you with everything you need to master mathematics, thermodynamics, fluid mechanics, and solid mechanics. Helping you to understand simple concepts and logic you need to solve any question.
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It's most especially built for people who could not afford quality education and for teachers who want to advance in knowledge.
We request you support, like, and share to reach out to those who need help.
You can make request on topic and write your doubts in the comment section. We will work effortlessly to get you what you need.
Please subscribe to this channel.
The Answer Is Golden | A Very Nice IIT JEE Math Problem |
In this video, I'll be showing you step by step on how to solve this IIT JEE Maths Exponential problem using a simple trick.
Please feel free to share your ideas in the comment section.
And if you are new here and you enjoy my content, please hit the like button and subscribe to my channel. Don't forget to hit the bell icon as well so you don't miss on my upcoming videos.
Please feel free to share your ideas in the comment section.
And if you are new here and you enjoy my content, please hit the like button and subscribe to my channel. Don't forget to hit the bell icon as well so you don't miss on my upcoming videos.
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Відео
How To Use The Lambert-W Function Like A Pro
Переглядів 20111 годин тому
In this video, I'll be showing you step by step on how to solve this problem using the Lambert W Function Please feel free to share your ideas in the comment section. And if you are new here and you enjoy my content, please hit the like button and subscribe to my channel. Don't forget to hit the bell icon as well so you don't miss on my upcoming videos. #education #school #learning #students #l...
India | Math Olympiad Trigonometry Problem | Find x
Переглядів 75613 годин тому
In this video, I'll be showing you step by step on how to solve this Olympiad Maths Trigonometric Problem using a simple trick. Please feel free to share your ideas in the comment section. And if you are new here and you enjoy my content, please hit the like button and subscribe to my channel. Don't forget to hit the bell icon as well so you don't miss on my upcoming videos. #education #school ...
A Very Nice Math Olympiad Problem | Solve for a
Переглядів 1 тис.18 годин тому
In this video, I'll be showing you step by step on how to solve this Olympiad Maths Exponential problem using a simple trick. Please feel free to share your ideas in the comment section. And if you are new here and you enjoy my content, please hit the like button and subscribe to my channel. Don't forget to hit the bell icon as well so you don't miss on my upcoming videos.
A Very Nice Math Olympiad Problem | Solve for x | A nice exponential simplification
Переглядів 76120 годин тому
In this video, I'll be showing you step by step on how to solve this Olympiad Maths Exponential problem using a simple trick. Please feel free to share your ideas in the comment section. And if you are new here and you enjoy my content, please hit the like button and subscribe to my channel. Don't forget to hit the bell icon as well so you don't miss on my upcoming videos.
A very nice maths olympiad | Can you solve?| Algebra problem
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In this video, I'll be showing you step by step on how to solve this Olympiad Maths Algebra problem using a simple trick. Please feel free to share your ideas in the comment section. And if you are new here and you enjoy my content, please hit the like button and subscribe to my channel. Don't forget to hit the bell icon as well so you don't miss on my upcoming videos.
Every nice maths olympiad algebra problem | Can you solve for all values of x | Algebra
Переглядів 420День тому
In this video, I'll be showing you step by step on how to solve this Olympiad Maths problem using a simple trick. Please feel free to share your ideas in the comment section. And if you are new here and you enjoy my content, please hit the like button and subscribe to my channel. Don't forget to hit the bell icon as well so you don't miss on my upcoming videos.
A very nice maths olympiad question | Solve for the value of x | Algebra problem
Переглядів 2,4 тис.День тому
In this video, I'll be showing you step by step on how to solve this Olympiad Maths Algebra problem using a simple trick. Please feel free to share your ideas in the comment section. And if you are new here and you enjoy my content, please hit the like button and subscribe to my channel. Don't forget to hit the bell icon as well so you don't miss on my upcoming videos.
Every nice maths olympiad algebra problem | Can you solve for a+b=? | Algebra
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In this video, I'll be showing you step by step on how to solve this Olympiad Maths Exponential problem using a simple trick. Please feel free to share your ideas in the comment section. And if you are new here and you enjoy my content, please hit the like button and subscribe to my channel. Don't forget to hit the bell icon as well so you don't miss on my upcoming videos.
A very nice maths olympiad question | Find the value of x | Algebra problem
Переглядів 45814 днів тому
In this video, I'll be showing you step by step on how to solve this Olympiad Maths Algebra problem using a simple trick. Please feel free to share your ideas in the comment section. And if you are new here and you enjoy my content, please hit the like button and subscribe to my channel. Don't forget to hit the bell icon as well so you don't miss on my upcoming videos.
A Very Nice Math Olympiad Problem | Solve for m and n
Переглядів 2,2 тис.14 днів тому
In this video, I'll be showing you step by step on how to solve this Olympiad Maths Exponential problem using a simple trick. Please feel free to share your ideas in the comment section. And if you are new here and you enjoy my content, please hit the like button and subscribe to my channel. Don't forget to hit the bell icon as well so you don't miss on my upcoming videos.
A Very Nice Math Olympiad Question | You Need To Know This Trick | Algebra
Переглядів 42214 днів тому
See the way I breakdown the solution of this question. There is a lot you can learn from this video. ENJOY If this is your first time to my channel, here, I shared simple step by step method of solving Algebra with a simple trick. Please like, subscribe, and share this video with your friends . Don't forget to comment if you have any questions or doubts or if you know a better way to solve this...
Germany || A Very Nice Math Olympiad Problem | Solve for the value of x
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In this video, I'll be showing you step by step on how to solve this Olympiad Maths Exponential problem using a simple trick. Please feel free to share your ideas in the comment section. And if you are new here and you enjoy my content, please hit the like button and subscribe to my channel. Don't forget to hit the bell icon as well so you don't miss on my upcoming videos.
Germany || Can You Solve This? || A Very Nice Math Olympiad Problem
Переглядів 47721 день тому
In this video, I'll be showing you step by step on how to solve this Exponential problem using a simple trick. Please feel free to share your ideas in the comment section. And if you are new here and you enjoy my content, please hit the like button and subscribe to my channel. Don't forget to hit the bell icon as well so you don't miss on my upcoming videos.
Harvard University Admission Question || A Nice Algebra Problem
Переглядів 1,6 тис.21 день тому
In this video, I'll be showing you step by step on how to solve this Radical problem using a simple trick. Please feel free to share your ideas in the comment section. And if you are new here and you enjoy my content, please hit the like button and subscribe to my channel. Don't forget to hit the bell icon as well so you don't miss on my upcoming videos.
A Very Nice Algebra Maths Problem || Radical
Переглядів 1,3 тис.21 день тому
A Very Nice Algebra Maths Problem || Radical
A very nice trigonometric problem || Mathematics || Algeria
Переглядів 45721 день тому
A very nice trigonometric problem || Mathematics || Algeria
A very nice olympiad maths question | x^2+x=y#-y, y^2+y=x^3-x | Can you solve this? | Algebra
Переглядів 291Місяць тому
A very nice olympiad maths question | x^2 x=y#-y, y^2 y=x^3-x | Can you solve this? | Algebra
A very nice olympiad question | How to solve a+2ab+b=8 Algebra |
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A very nice olympiad question | How to solve a 2ab b=8 Algebra |
A very nice maths olympiad question | Functional equation |
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A very nice maths olympiad question | Functional equation |
A very nice olympiad question | This question is tricky! | Algebra |
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A very nice olympiad question | This question is tricky! | Algebra |
Poland | A very nice maths algebra probem
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Poland | A very nice maths algebra probem
International Maths Olympiad Problem | Solve for x,y, and z | Algebra
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International Maths Olympiad Problem | Solve for x,y, and z | Algebra
A very nice Maths Olympiad question | How to solve 1/m + 4/n = 1/12 | Algebra
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A very nice Maths Olympiad question | How to solve 1/m 4/n = 1/12 | Algebra
USA | A Very Nice Exponential Maths Olympiad Question | The Only Trick You Need!
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USA | A Very Nice Exponential Maths Olympiad Question | The Only Trick You Need!
USA | A Very Nice Maths Olympiad Question | The Only Trick You Need!
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USA | A Very Nice Maths Olympiad Question | The Only Trick You Need!
International Maths Olympiad Problem | Solve for all tripples (x,y,z) | Algebra
Переглядів 1,7 тис.Місяць тому
International Maths Olympiad Problem | Solve for all tripples (x,y,z) | Algebra
International Maths Olympiad Problem | How to Solve n^6-n^3=2 | Solve for real solutions
Переглядів 1 тис.Місяць тому
International Maths Olympiad Problem | How to Solve n^6-n^3=2 | Solve for real solutions
A very nice olympiad question | You need to try and solve this | Algebra |
Переглядів 816Місяць тому
A very nice olympiad question | You need to try and solve this | Algebra |
Okay
3^(×-1)=5^(1-×^2) =>3=5^(1+×) =>×={(log @5(3)-1}
I don't want to be cocky but this peoblem us way too easy,for Olympiad
Quite a easy one
Too much talking in very easy places
We can solve it, so the answer is likely to be nice. Root 15~4, so approx 8^x+0=62. Thus x is about 2. Check 2. The middle trrms cancel, giving 16+15+15+15=62.
5³ᵃ = 10 3a · lg(5) = 1 a = 1 / [3 · lg(5)] a ≈ 0,477 (which is very close to lg(3), by coincidence)
😂(×+y)(×-y)=98/2=49 49=49(1)or 7(7)(n.a.) ×+y=49&×-y=1 =>×=25&y=24
Isn't the answer 1.
a=ln(10)/(3ln(5)). This is pretty easy to generalize.
Why ln? If you use lg instead, you get a = 1 / (3 · lg(5)).
Yah he is making it more complex.
{(x^3)^2-[(x-1)^3]^2}=0 [x^3-(x-1)^3][x^3+(x-1)^3]=0 {(1)[x^2+x(x-1)+(x-1)^2]}{(2x-1)[x^2-x(x-1)+(x-1)^2]}=0
62=16+15+16+15=16+15+8*15+16+15-8*15=4^2+(*15)^2+2×4×*15+4^2+(*15)^2-2×4×*15=(4+*15)^2+(4-*15)^2 (4+*15)^x+(4-*15)^x=(4+*15)^2+(4-*15)^2 (4+*15)^x=(4+*15)^2buradan x=2 (4-*15)^x=(4-*15)^2 buradan da x=2. Təşəkkürlər.
Excellent delivery 👏
The equation reduces down to (x² + 1 + 1/x²) / (x + 1 + 1/x) = 3 Let u = x + 1 + 1/x u² = (x + 1 + 1/x)² = (x² + 1 + 1/x²) + 2(x + 1 + 1/x) = (x² + 1 + 1/x²) + 2u u² - 2u = (x² + 1 + 1/x²) The equation (x² + 1 + 1/x²) / (x + 1 + 1/x) = 3 becomes (u² - 2u) / u = 3 u = 5 x + 1 + 1/x = 5 x² - 4x + 1 = 0 x = 2 ± √3
Superb! 👏
رائع
Approximation method,x=2 insertion can be easy.But if the question x=more than 10,you need more difficult tackling.
12:23
(4 + roo15) ² + (4-root15)²=2(16+15)=62 so x=2
Why is x-3=0?
How did u get the coefficients for the quartic? 1-4-6-4-1?
It appears on both sides. You can also move all the terms to one side of the equation and factor out x-3, and you'll still be having the same result.
@@willemesterhuyse2547 it was set equal to zero. I believe it's called the zero-property rule? Have you not heard of this? Do u know how to factor?
@@shannonmcdonald7584 One may also choose to cancel x-3 on both sides and then you won't find the x=3 solution. This is problematic since one may test the x=3 solution directly by substitution. I know how to factor.
Verrrry verrrty good tank you🙏🙏🙏
you can totally recognise the pattern which is 5! = 120 then the value of x is undeniably 1, so easy
Alternatively, the second integer solution of x=-6 can also be easily "seen" from the original form of the equation in much the same way that we spotted x=1. If you write the 120 = 5! = (-5)*(-4)*(-3)*(-2) = (x+1)(x+2)(x+3)(x+4), then you will immediately recognize that x+1 = -5 will satisfy it. Thus x=-6 is a solution.
It's a quartic equation with real coefficients (x^4 + 10x^3 + 35x^2 + 50x - 96 = 0). It has two integer solutions and two complex-conjugate solutions. The two integer solutions can be easily guessed. The first one (x=1) can be guessed from the original form of the equation, the second solution (x=-6) can be guessed after splitting off the linear factor (x - 1) and testing the integer divisors of the last coefficient of the resulting cubic equation (x^3 + 11x^2 + 46x + 96 = 0). After splitting off another linear factor (x + 6), we arrive at a quadratic equation (x^2 + 5x + 16 = 0). That one can be solved by the usual formula. It doesn't have any real solutions but we get the two complex-conjugate solutions. They both have a real part of -5/2 and an imaginary part with an absolute value of 1/2*sqrt(39). Easy.
Wonderful! 👏
1 and -6 are easy solutions to get by asking what four numbers in a sequence increasing by 1 are the same as 2 * 3 * 4 * 5. you can see 1 will give you this sequence from this problem. And notice that if all the numbers are negative you get the same solution so -5 * -4 * -3 * -2 this sequence can be found by putting in -6. The other 2 solutions appear to be imaginary or complex from me quickly looking at it. We know that the angles of the 4 consequent numbers must be added together to make the overall angle 0. the issue is the angle will change between the four numbers. Meaning no more eyeballing the problem
Nice one! 👏
I just looked at the prime factorization of 120 (2^3*3*5) and realized u can use this for x=1(2*3*4*5) or - 6 (u can put a - sign on each of 2,3,4,5 bc they r an even number and it cancels out). Then I multiplied the given expression out (x^4+10x^3+35x^2+50x+24-120=0) and did a polynomial long division with the one given by the 2 found solutions ( (x-1)*(x+6) = x^2+5x-6), which gives x^2+5x+16. With the quadr. form. we get (- 5 +/- sq(39)i)/2 for the other 2 solutions.
Excellent! 👏
1? 120 can be decomposed into a multiplication of primes. : 2 × 2 × 2 × 3 × 5 .. so it was a guess.
Yeah. But that's not the only solution.
Good one
Thank you! Cheers!
this equation was resolved by substitution from a 4th degree power to quadratic equation via k, is there a way we can do this for cubic equation?
In this case it worked because the coefficients of x^2 and x term were the same
That technique only worked in this particular case due to the specific values of the coefficients. It's not possible to do this in the general case. For (depressed) cubic equations there are some other general ways to reduce them to a quadratic equation in some auxilliary variables from which the general solution to the cubic can be constructed (Cardano's method). But that method is more complicated; there's a whole theory for it.
Very intelligent
If we add both entities like m raised to 2 -n and later term we can get m raised 2- m + n raised 2-n=98 By hit and trial we can satisfied above equation with takin value 7 and 8 for m and n
useful formula x^3 + y^3 = (x + y)^3 - 3xy(x + y) ax^2 + bx + c = 0 and b is even(b = 2b') => x = (-b' ± √(b'^2 - ac))/a
My approach is: (a+b)(a-b)=4 (a^2+b^2+4)(a^2+b^2-4)=16 (a^2+b^2)^2=32 a^2+b^2=4root(2) (notice that a^2 and b^2 are both positive, so no plus or minus sign is needed for this square root) (a+b)^2=4root(2)+4 a+b=plus or minus 2root((root(2)+1))
i got a+b = root( root(32)+4 ), is it correct or where did i do it wrong
oh it is the same thing, i just didnt simplify it, i am not very good at math haha
√32=4√2 So √(4+√32)=√(4+4√2) = 2√(1+√2) Therefore they are equal
I see, thank you
I am bored so lets solve with complex numbers. 4+4i=a²+2abi-b²=(a+bi)² a+bi=√(4+4i) Let z=4+4i |z|=4√2 z/|z|= 1/√2 + i/√2 Arg(z)=π/4 Therefore √z=2∜2 * e^(πi/8) Which as shown earlier is equal to a+bi. a=Re(√z), b=Im(√z) a=2∜2 * sin(π/8) b=2∜2 * cos(π/8) a+b=2∜2 * √2 sin(⅜π) a+b= 2^(¾) * √(2+√2) а+b=2√(1+√2) Naturally while taking the square root of z instead of Arg(√z)=⅛π we can use Arg(√z)=⅝π which will by similar methods turn out to be the other solution in which a+b=-2√(1+√2)
A Very Nice Math Olympiad Problem: x³ + x² = 4/27; x =? x³ + x² - 4/27 = 0; (x³ - 1/27) + (x² - 3/27) = [x³ - (1/3)³] + [x² - (1/3)²] = 0 Let: a = 1/3; [x³ - (1/3)³] + [x² - (1/3)²] = (x³ - a³) + (x² - a²) = 0 (x - a)(x² + ax + a²) + (x - a)(x + a) = (x - a)(x² + ax + a² + x + a) = 0 x - a = 0, x = a or x² + ax + a² + x + a = x² + 4x/3 + 4/9 = (x + 2/3)² = 0, x + 2/3 = 0 x = 1/3 or x = - 2/3; Double root Answer check: x = 1/3: x³ + x² = 1/27 + 1/9 = 4/27; Confirmed x = - 2/3: - 8/27 + 4/9 = 4/27; Confirmed Final answer: x = 1/3 or x = - 2/3; Double root
This is fantastic 👏
√15≈√16 so: 4−√15≈0 and 4+√15≈8 0²+8²=64 and 64≈62 Let's try x=2 (4+√15)²+(4−√15)²=(16+8√15+15)+(16−8√15+15) (4+√15)²+(4−√15)²=32+30=62 x=2 is indeed a solution of f(x)=(4+√15)ˣ+(4−√15)ˣ−62 f(x)=nˣ+1/nˣ−k is symmetrical with respect to the origin and has 2 solutions as a sum of two exponentials, one increasing and the other decreasing. If x is a solution of f(x)=nˣ+1/nˣ−k then -x is the other solution of f(x)=nˣ+1/nˣ−k Here we have n=4+√15 and 1/n=4−√15 because (4+√15)(4−√15)=4²−(√15)²=1 f(x)=nˣ+1/nˣ−62 having solution x=2 so x=-2 is also a solution.
Excellent!
Where did you come up with -2 as a root of the cubic, though, lucky guess?
The factor theorem
👍👍
I love how you did both a log version and a pure exponent version. The exponent version required the 6 to show up where it did, tho. You couldn't use that method if there were different numbers/ bases on each side
I don’t see how you concluded a-2 and a-3 were factors except by intuiting that 97 is the sum of two numbers which are 2 perfect quartic roots. And since you can restate the problem as: find all x,y such that x,y are natural numbers, x+y=97, and x^.25 + y^.25 = 5, then brute force is faster than the binomial expansion, 2 polynomial divisions, and one quadratic solve. Even when replacing 97 with 195857 and 5 with 34, brute force will be faster.
Factor when ever you can. Good idea to start solving.
Love the math you showed. I solved it just be realizing that they have to be integers and you need to end on an integer so they have to be perfect squares as well as positive (or you'd have an i on the RHS), so just cycle through them starting with 1, 4, 9, and there's your answer: 4, 9
Yeah, you're right.
10:28 I would have reduced the 2 from the bottom before I split into 2 cases.
Yeah, I agree with you 💯
Really excellent, although to be fair the original was contrived to be able to simplify this far. Still loved your work. I think I would have gone from (root 9)^2 => 9 a little quicker and once I had shown the trick went from (root 9)^2 - (root 5)^2 => 9 - 5 = 4. But that's just a style choice
This one was easy i got the ans under 2 min !!! thank you for the question !
Well done!
Wonderful job. You could have given the approximate values which are close to -7.5 and +6.5.
Hey can anyone tell me what i am doing wrong here , x^6 = x^4 + x^2 x^6 - x^4 - x^2 = 0 x^3 - x^2 - 1 = 0 x ( x^2 - 1 ) = 1 x cant be 1 as it will form a equilateral triangle and we know angle given is 90 so x^2 - 1 = 1 so x^2 = 2 and x = root2 and x = root2 doesnt satisfy (x^3)^2 = (x^2)^2 + x^2
You did an excellent job. Check line three of your solving. That's where you made a mistake.
@@SpencersAcademy Ohh i made the mistake there understood thank you , appreciate it 😊
Interestingly, Desmos thinks there is only one solution to the equation
this question looked so simple on the first glance , great explanation and wonderful question choice
Glad it was helpful!
X^3 + X^2 - 4/27 = 0 --- (1) X = 1/3 ( by trial and error ) divide eq (1) by (x - 1/3) get the other value of x as -2/3 and -2/3 ( twice ) so finally three values of x are 1/3 , -2/3 , -2/3 Edit - personally i think this is a better way and quicker way , let me know if im doing something wrong
Fantastic
p^2-62p+1=0 çevrilmiş kvadrat tənlik olduğundan p=62/2 +*(62/2)^2-1=31+*31^2-1=31+*961-1=31+*960=31+*960=31+*64×15=31+8*15=(4+*15)^2; p=31-8*15=(4-*15)^2 (* kökaltı kimi başa düşülsün) kimi həll etsə idiniz kiçik ədədlərlə işləməli olardınız ki, bu da Sizin əməyinizi yüngülləşdirərdi. Təşəkkürlər. (* kökaltı