Great video with an elegant solution! I was able to solve this in my head (without writing things down), before watching the video, just wanted to share how I did it (slightly less elegantly) Actually that path on the scratch paper can work, just need to notice some possibilities for simplification (the ones you mentioned in the video). My first observation was the if x1 is a real solution, then -x1 is also a different one, so it's enough to focus on finding two different positive solutions. Then I just used the quadratic formula for x^2 (I tend to avoid introducing new letters if the original "expression" is simple enough to carry over, so instead of "u" in the video, I just use "x^2" everywhere): (x^2)1,2 = (40±√(1600-4q))/2. Observation: this can be simplified by making the division: (x^2)1,2 = 20±√(400-q). Observation: q has to be less than 400 to have two values for x^2 (which enables 4 for x). These solutions can be used to express the original equation with a multiplication of two polynomials: (x^2-(20+√(400-q)))*(x^2-(20-√(400-q)))=0. From here, using the (a+b)(a-b)=a^2-b^2 formula, this can be expressed as: (x+√(20+√(400-q)))*(x-√(20+√(400-q)))*(x+√(20-√(400-q)))*(x-√(20-√(400-q)))=0. From this, it's actually possible to derive the four roots in ascending order: -√(20+√(400-q)), -√(20-√(400-q)), √(20-√(400-q)), √(20+√(400-q)). Observation, for these to be real and different, 20-√(400-q) has to be larger than 0, which means q has to be greater than 0. Let d be the positive real which denotes the increment between these solutions. Then from the two middle solutions we know that -√(20-√(400-q)) + d = √(20-√(400-q)). So d = 2*√(20-√(400-q)). From the two other solutions we also know that -√(20+√(400-q)) + 3d = √(20+√(400-q)), so 3d = 2*√(20+√(400-q)). At this point I recognized that if the square root is removed, adding these together would make q go away. So square both: d^2 = 4*(20-√(400-q)) and 9d^2 = 4*(20+√(400-q)), and add the two equation together: 10d^2 = 4*40. So d = 4 (note that d was a positive real; but if we'd enable -2, that would mean that for the same q value we'd find the same solutions, just in descending order). Substituting d into one of the previous equations: 4 = 2*√(20-√(400-q)), which gives only a single possible value for q: 144 (which was the question). With that q value (as a check), the roots are -6, -2, 2, 6.
blackpenredpen We can do it in a much easier way called RH criteria (control systems) auxiliary equations solution we will get and roots are symmetrical about origin ...if u need solutions just ask me.
@blackpenredpen just consider A.P of a-3d, a-d, a+d, a+3d then sum of all the roots is zero hence a=0 then solve for d by multiplying two roots at a time that gives d=2,-2 hence -6,-2,2,6 multiply all to get q =144
Another way of solving the question is as follows: You can assume the roots of the original equation to be: (a-3d),(a-d),(a+d),(a+3d) where 'a' is the first term of A.P. and '2d' is the common difference. Further, you can solve for 'd' by 'Vieta's relations'. BTW, LOTS OF LOVE FROM INDIA!
@@36sufchan Yes. But since the coefficient of x^3 is 0 (that comes from the Vieta formula that was referred to earlier), the sum of the roots is 0, and so (a+3d)+(a+d)+(a-d)+(a-3d)=0, implying a=0.
This is the way I solved it: At first I also assumed that u is equal to x^2, which gives you the following equation. u^2-40u+q=0 If you then solve for u, you get this equation. u=20+/-sqrt(400-q) Now I used the fact, that x is the square root of u, which means, that the solutions for x aren't just a,b,c and d but -b, -a, a and b. Because the solutions have to form an A. P. b-a has to be equal to a-(-a): b-a=a-(-a) b-a=2a b=3a Now we know, that the four solutions for x are -3a, -a, a and 3a. The solutions for u have to be the squares of these four numbers, because I assumed, that x^2 is equal to u. If you square the four numbers you get a^2 and 9a^2. If we now come back to the formula for u, we know that 20+/-sqrt(400-q) is either a^2 or 9a^2 depending on whether you add or subtract sqrt(400-q). Because a^2 is positive you get 9a^2 if you add and a^2 if you subtract. Now you have two equations: 20+sqrt(400-q)=9a^2; 20-sqrt(400-q)=a^2 Now you can multiply the right equation by 9 to get 180-9sqrt(400-q)=9a^2. Then you can combine this equation with the left equation to get the following. 20+sqrt(400-q)=180-9sqrt(400-q) Now you can finally solve for q: sqrt(400-q)=160-9sqrt(400-q) 10sqrt(400-q)=160 sqrt(400-q)=16 400-q=256 144-q=0 144=q
what i did before watching: let the roots be a, a+b, a+2b, a+3b then * can be written as (x-a)(x-a-b)(x-a-2b)(x-a-3b) by comparing coefficient of x^3, we get (-4a-6b)=0, b=-2a/3 so * can be re-written as (x-a)(x-a/3)(x+a/3)(x+a) now compare the coefficient of x^2, we get a^2(-10/9) = -40, a=6 therefore, * is (x-6)(x-2)(x+2)(x+6) q = (-6)(-2)(2)(6) = 144
Yes I did the same. The question is very straight forward and took 1min to solve. Then I saw he has solved in 10 min and thought my ans is wrong. But fortunately I am current.
I solved it as follows: The function is even so if r is the smallest positive root then -r is as well. Also 3r and -3r are roots (in order to form AP). It follows that: Eq1: 81r^4 -360r^2 +q=0 Eq2: r^4 -40r^2 +q=0 (Eq1)-(Eq2) implies 80r^2 (r-2)(r+2)=0, Thus r=2 (positive) So our original equation of the question becomes (x+6)(x+2)(x-2)(x-6)=0 Now it's clear that the constant term is 144 (which is q)
My first step was similar. But I calculated coefficient at x^2, knowing by Viet fotmula that it is a sum of all products of two roots. It equals -10*r^2 = -40, so r=2, so I know all roots and just multiplied them all by Viet again
Start with the sum. Since sum of roots is zero, the roots would have to be -3d, -d, d, 3d Then, sum of pairwise products is -40, so -d^2 + -9*d^2 = -40 so d = +2, -2 Roots: -6, -2, 2, 6 q = (-1)^(degree of polynomial) * product of all roots. = (-1)^4 * (-6*-2*2*6) = 144
I like the solution here exploiting the symmetry, but this can also be done by brute force algebraic manipulation easily enough: Solns a,a+b,a+2b,a+3b. Product(x-a-kb)=x⁴-40x²+q=0. Compare x³ terms: 4a+6b=0 => b = -⅔a. Compare x² terms: 6a² + 18ab + 11b² = -40. => -10/9 a² = -40 => a = ±6, b = -±4. Compare x⁰ terms: q = (-6).(-2).2.6 = 144.
ok so let's try to solve it before i watch: t = x^2 t^2 - 40t + q = 0 for 2 different solutions (-40)^2 - 4*1*q > 0 1600 - 4q > 0 q < 400 ----------------------- for both to be positive from Viete's formulas -(-40) / 1 > 0 (their sum) and q / 1 > 0 (their product) So my guess is 0 < q < 400
i also did this, i dont think its wrong (cos 144 is indeed between 0 and 400) but it just isnt complete because not all values of q between 0 and 400 give you arithmetic progression when you sub it into the original eqn
Easy way a - 3d , a - d , a + d , a + 3d in ap with c.d = 2d Sum of roots = 0 4a = 0 a = 0 Taking products 2 at a time total combimations - 4 C 2 = 6 - 3d × - d + - 3d × d + - 3d×3d + - d×d - d×3d + 3d × d = - 40 - 10d^2 = - 40 d = +- 2 Nos. are - 6 , - 2 , 2, 6 or vice versa
I just made q=mn, and then made the equation into (x^2-m)(x^2-n)=0, then solved that to get +-sqrtm and +-sqrtn. Then I just did sqrtm-sqrtn = sqrtn-(-sqrtn). This gave m = 9n, plug that into m+n=40, get n=4, and m=36, so q=144. This solution took about 2 minutes to solve. Also, whether n > m or m > n doesn't affect the solution, as either way one of m or n will equal 9n or 9m.
I don't even use this kind of math in my classes anymore, I just watch the videos can you make them fun and I also love math, but mainly because of you.
If you start with a polynomial (x-a)(x-b)(x-c)(x-d) you have a+b+c+d=0 since it is an arithmetic progression, you can then prove that the roots have the form a, a/3, -a/3 and -a so the polynomial becomes (x2-a2)(x2-a2/9) and you get a2 = 36 and a4/9 = 144 = q
I found it easier to do Vieta on the original quartic with roots a1, a2, a3, a4 with a1=A, a2=A+r, a3=A+2r, a4=A+3r. Then 4A+6r=0 (vieta rules, coefficient of x^3) => A=-3/2r WLOG, a1 r=4 (since we assumed r>0, a
Alternative method: Take the four solutions as a-3d, a-d, a+d, a+3d. Then sum of roots = 4a= -(coefficient of x^3/coefficient of x^4)= 0, which implies a=0. Then the product of roots taking two at a time, and summing it up is given by coefficient of x^2/coefficient of x^4, which leads us to the equation -10d^2 = -40, which gives d=±2. Then we have the solutions as -6, -2, 2, 6. Constant term/coefficient of x^4 = product of roots implies, q=-6*-2*6*2=144
a-3d, a-d, a+d, a+3d be 4 roots. Step 1 compare coeff of x^3 with sum of roots Result a=0 Step 2 compare coeff of x^2 with sum of product of roots taken 2 at a time Result d=2 Step 3 compare coeff of x^0 (or constant term) with product of all roots Result q=144
Do you know Michael Penn??? He's keep saying great like what you did on this vid. Anyway he's great in math as well. I followed hin here in yt just like you. You both are great. Thanks Sir Steve.
There are many solutions for the equation itself, but only one q works for constant spacing when sorted; the rest have integers and radicals. Under 400, q=144 delivers all integer solutions not just sequential.
Actually, once you find r1+r2 = 40, you know they must be the sum of two squares. It is quite easy to see by inspection that only 4 and 36 fit. So the solutions must be +- 2 and +- 6. The sequence -6, -2, +2,+6 is arithmetic with d=4.
I did it pretty much the same way. Towards the end looked different because I noticed that the square root of r2 equals 3 times the square root of r1, and so I substituted it into the r1+r2=40 formula directly, which gives you that 10r1=40, so r1=4, and r2=9r1 (or 41-r1)=36. I was effectively using the idea of “d” without it appearing on the paper. I wish you hadn’t said the solution at the beginning. At first I had to train my brain not to think of factor pairs for 144.
The solutions could be generalised, If say the equation was x^4 - a x^2 + q = 0 and the AP difference of the solutions is d, then for all n, {d = 2n, a = 10n^2, q = 9n^4 }
If an easier solution that gives ALL the values for q : Let u=x², we get u²-40u+q=0. If we want 4 real solutions, we need 2 real solutions for u that will give 4 real solutions for x. In order to get 2 real solutions for u, we need d = (-40)² - 4q > 0 *q < 400* Therefore the answers are u = [40 +/- sqrt(d)] / 2. As u=x², then, we get x = +/- sqrt(u). As u can take two real values, then x can take 4 real values if u>0 (the sqrt function is defined on [0,+infinity[). This, means that : [40 +/- sqrt(d)] / 2 > 0 and after some calculations, we get *q > 0* . Therefore, x^4 - 40x² + q = 0 has four real solutions when 0 < q < 400. We can notice that q=144 is between 0 and 400.
Simply take the roots as a-3d,a-d, a+d,a+3 d, the 4 roots in AP. Sum of the roots is zero, makes a=0 and the sum of their 6 products taken 2 at times equal to 40 solves the roots.
My solution: First I wrote the roots as(solved the quadratic equation for x^2) x1,2,3,4=+-sqrt(20+-sqrt(40^2-4q)/2) Then I used Newton's method to solve the following system of nonlinear equations for (q,a,d): x1=a or x1(q)-a=0 x2=a-d or x2(q)-a+d=0 x3=a-2d or x3(q)-a+2d=0 d is the common difference of the arithmetic progreasion.
My thought process: 1. Observe that the sum of four roots are zero because the coefficient of x^3 is zero. Let the roots be -3k, -k, k and 3k such that they form an ap with common difference 2k. Note that q = the products of roots = 9*k^4 2. Substitute y = x^2, the equation becomes a quadratic equation of y. By symmetry, the two roots of y corresponds to (3k)^2 = 9k^2 and k^2. As for each solution of y, there will be two values of x, differ by +/- sign. 3. The sum of root of the quadratic equation of y is 40. Therefore, 9*k^2 + k^2 = 10*k^2 = 40 ==> k = +/- 2. But the sign does not matter here as, again, because of symmetry. 4. q = 9*2^4 = 144 and the roots are -6, -2, 2 and 6. QED =]
I just factorize it to (x²-a)(x²-b). Then, by guessing, with a+b=40, I just put in 4 and 36 as they are square numbers. Then it turns out that it works
i just wrote a general solution that would form an arithmetic sequence in the form (x^2-u^2)(x^2-(3u)^2) where u is the positive root with smallest magnitude (roots are -3u, -u, u, 3u, 2u between all of them) expanded this out to get x^4-10u^2x^2+9u^4 the x^2 term is also -40 so 10u^2=40 => u=+ - 2 plus or minus doesn't really matter since its symmetrical so just gonna use +2 sub this into 9u^4 get 9x16=144 so q = 144 and the ending quartic is x^4 - 40x^2 + 144
hey blackpenredpen, there's been a question ive been struggling on and i hope you can help me solve it. The question in full is: 15*(5/4)^(n-1) > 5n+10 and solve for the conditions for n. This question is related to the geometric and arithmetic nth terms, I skipped ahead and made the conditions for both sides. Please have a look at this for me
Can someone explain me (in a way that a common mortal can understand) how do we know that the difference between the sqrts of solutions is "d" and it is costant?
That hides in "the solutions follow an arithmetic progression", which means that the difference between two numbers, if written in ascending order, is always the same.
Hi! I've been watching your videos for weeks and I am really fascinated. I love math and I really want to study it at advanced level and be good at it, but I don't know how to begin. Have you got any advices(maybe some videos or sites or books)? I hope you read this message Thank you, you're amazing
When I paused the video, I kinda stumbled across the solution by accident, before I started doing any real maths: I saw that the graph would be symmetrical, and would look like a W. So just to get an idea of the shape, I set q=0, and worked out the values from x=-3 to x=3. I noticed that -2 and +2 both gave -144. On a whim, I tried it for 6, and sure enough, -144. So then it was obvious that for q=144, you'd get solutions for x=0 at -6, -2, 2, and 6. I was kinda disappointed - I feel like the Olympiad writers should have tweaked the equation a little, put different coefficients etc, to make the solution less easy to just stumble across.
Here's my go at it: To get 4 real sols, the quadratic with z = x^2 needs two positive sols b > a > 0, so the 4 real sols are +/- sqrt(a) and +/- sqrt(b). Let r^2 = a and s^2 = b. We know the order is -s, -r, r, s, so the gap is r-(-r) = 2r and so s = r + 2r = 3r. That means the solutions to the quadratic are a and 9a. (z - a)(z - 9a) = z^2 - 10az + 9a^2 = z^2 - 40z + q, so a = 4 (and b = 36) and q = 144. Solutions are -6, -2, 2, 6.
BlackpenRedpen, I've watched your 'Integrating sec^n (x)' video. I actually have an interesting issue with it.. *Please look at this, people* I derived the values of Integral of sec^2(X), sec^3(X) , ... till the 6th power of sec (X). On the other hand, I verified the answers and differentiated them to check if they were right. The issue is with the answer of Integral of sec^4 (X). I derived the value of this problem with 2 different steps. Int. sec^4 (X) = Int. sec^2(X) * sec^2 (X) dx. I used the DI Method to work this out. The answer will be, sec^2(X).tan(X) - 2 Int.(sec^2(X)*tan^2(X)*). 1st method: I did 'U- Substitution' for tan(X) and ended up with, sec^2 (X).tan(X) - 2 Int.( u^2. du) = sec^2(X).tan(X) -2/3* tan^3(X)+ c. When I differentiate this, I get sec^4(X), which is all perfect.. 2nd method: Instead of doing 'U-Substitution' for tan(X), I replaced tan^2(X) with sec^2(X)-1. And when I did the following, I ended up with the answer, 1/3*sec^2(X).tan(X) + 2/3*tan(X) + c *which is also the answer derived by using the Int.sec^n(X) formula*.. When I differentiate this I end up with, 7/3*sec^2(X).tan^2(X), instead of sec^4(X).. Please help me with this. It took me really long to type this perfectly. It'll be worth it if I get a good response. Had to comment in this video, because it's the Latest one. Please help me. Thanks!
(1) 1:36 . . . Curious - Could complex numbers not form APs as well? (2) 2:16 . . . If all the roots are equal they also form an AP - a trivial case AP with zero common difference.
Dear all, In this lockdown stage in home, please provide game equipments to your children/students to play. If not help them to watch "Math Art Studio" in you tube. They will play with their names and learn different concepts in mathematics.Those who have seen it they have learnt maths and enjoyed its beauty every day.
So if you assign x=2 and q=144 to the original expression it actually equals 0. But if you try with different values for x it doesn't work out anymore.
@@blackpenredpen I meant: x^4 -40x^2 + q = 0 Assigning the value 2 to x and the value 144 to q you actually make it work the result is 0. But if you assign a different value to x like for example 3 it doesn't work anymore, it doesn't give you 0 that expression. So basically you solved for q but only for the case x=+-2 ?
@blackpenredpen plz some tough Indian Olympiad problems basically RMO ( Regional Mathematical Olympiad) and INMO ( Indian National Mathematical Olympiad)... I would like them albeit you solve a few IIT JEE ADVANCED problems. Plz some as many integrals of IIT and some Indian Olympiad problems to help us Indians wo sometimes are unable to understand the solution. Thank you Regards Abhijeet
Here is the Nike way : x=a , a+c , a+2c , a+3c then a^4 - 40a^2=(a+c)^4 - 40(a+c)^2=(a+2c)^4 - 40(a+2c)^2=(a+3c)^4 - 40(a+3c)^2 after doing a little bit of algebra you will get 3 equations with coefficients : 4 6 4 1 -80 -40 8 24 32 16 -160 -160 12 54 108 81 -240 -360 make sure to put them in the same order then simplify to get 4 6 4 1 -80 -40 0 -12 -24 -14 0 80 0 0 24 36 0 0 the last one tells us that: 24ac^3+36c^4=0 --> 12c^3(2a+3c)=0 a=-3/2c since c cannot be 0 plug in the first equation to get c=+-4 -->a=-+6 don't care which one is just plug any a into the original equation 6^4 - 40(6)^2 +q =0 q=144 first I made a mistake (forgot to square the 3 after plugging a=-3/2c) but then fixed it
Well, some ppl might have tried the problem already so I just wanted to tell them the numerical answer right away. On top of that, the numerical answer is never that interesting, but the solution is. : )
❤ suppose that roots are a - 3d , a - d , a+d , a+3d since coefficient of x^3 = 0 , sum of roots = 0 or a = 0 now roots are - 3d , - d , d , 3d equation will be (x^2 - 9d^2) (x^2 - d^2) = 0 or x^4 - 10 d^2 x^2+9d^4 = 0 comparing coefficient of x^2 and constant term 10d^2 = 40 , 9d^4 = q d^2 = 4 hence q = 9× 16 = 144
My scratch paper: (don’t do this) twitter.com/blackpenredpen/status/1260624888735268864?s=21
Bro I think that the solution would have been done in less steps
Just take the roots to be a-3d,a-d ,a+d,a+3d
How would you integrate 1/((e^x)-x))
Great video with an elegant solution! I was able to solve this in my head (without writing things down), before watching the video, just wanted to share how I did it (slightly less elegantly) Actually that path on the scratch paper can work, just need to notice some possibilities for simplification (the ones you mentioned in the video). My first observation was the if x1 is a real solution, then -x1 is also a different one, so it's enough to focus on finding two different positive solutions. Then I just used the quadratic formula for x^2 (I tend to avoid introducing new letters if the original "expression" is simple enough to carry over, so instead of "u" in the video, I just use "x^2" everywhere): (x^2)1,2 = (40±√(1600-4q))/2. Observation: this can be simplified by making the division: (x^2)1,2 = 20±√(400-q). Observation: q has to be less than 400 to have two values for x^2 (which enables 4 for x). These solutions can be used to express the original equation with a multiplication of two polynomials: (x^2-(20+√(400-q)))*(x^2-(20-√(400-q)))=0. From here, using the (a+b)(a-b)=a^2-b^2 formula, this can be expressed as: (x+√(20+√(400-q)))*(x-√(20+√(400-q)))*(x+√(20-√(400-q)))*(x-√(20-√(400-q)))=0. From this, it's actually possible to derive the four roots in ascending order: -√(20+√(400-q)), -√(20-√(400-q)), √(20-√(400-q)), √(20+√(400-q)). Observation, for these to be real and different, 20-√(400-q) has to be larger than 0, which means q has to be greater than 0. Let d be the positive real which denotes the increment between these solutions. Then from the two middle solutions we know that -√(20-√(400-q)) + d = √(20-√(400-q)). So d = 2*√(20-√(400-q)). From the two other solutions we also know that -√(20+√(400-q)) + 3d = √(20+√(400-q)), so 3d = 2*√(20+√(400-q)). At this point I recognized that if the square root is removed, adding these together would make q go away. So square both: d^2 = 4*(20-√(400-q)) and 9d^2 = 4*(20+√(400-q)), and add the two equation together: 10d^2 = 4*40. So d = 4 (note that d was a positive real; but if we'd enable -2, that would mean that for the same q value we'd find the same solutions, just in descending order). Substituting d into one of the previous equations: 4 = 2*√(20-√(400-q)), which gives only a single possible value for q: 144 (which was the question). With that q value (as a check), the roots are -6, -2, 2, 6.
came for math, stayed for pikachu
-6,-2,2,6 arithmetic progression
blackpenredpen We can do it in a much easier way called RH criteria (control systems) auxiliary equations solution we will get and roots are symmetrical about origin ...if u need solutions just ask me.
I am your fan but can you suggest me how can I get advance calculus explore book
0,0,0,0 matters?
@blackpenredpen just consider A.P of a-3d, a-d, a+d, a+3d then sum of all the roots is zero hence a=0 then solve for d by multiplying two roots at a time that gives d=2,-2 hence -6,-2,2,6 multiply all to get q =144
@@gz4978 it is not arithmetic progression
My solution was: "They were probably nice giving whole number roots, what two perfect squares add up to 40?"
That's even quicker 👏🏼
I wonder what the maths professor would think of this approach 🤔
Mine too
Social engineering
My grandmother does stuff like this xD. However, I'm more stubborn haha.
Another way of solving the question is as follows:
You can assume the roots of the original equation to be:
(a-3d),(a-d),(a+d),(a+3d) where 'a' is the first term of A.P. and '2d' is the common difference.
Further, you can solve for 'd' by 'Vieta's relations'.
BTW, LOTS OF LOVE FROM INDIA!
This is what I did. The roots sum to 0 so a=0, then Vieta on the second coefficient gives -10r=-20 so r=2, making the sesquence -6,-2,2,6.
Won't the first term of the AP be a-3d?
@@ProfOmarMath yupp precisely
@@36sufchan Yes. But since the coefficient of x^3 is 0 (that comes from the Vieta formula that was referred to earlier), the sum of the roots is 0, and so (a+3d)+(a+d)+(a-d)+(a-3d)=0, implying a=0.
@@ProfOmarMath I see. Thanks for the explanation!
when you realise the pikachu is a microphone
*_biggest maths anime betrayal in history_*
Nah. It's just clipped to its back. (10:08)
blackpenredpen in quarantine day 2000: Speaking to a Pikachu
What will I do in day 300?
@@blackpenredpen you will watch pikachu do the math
Btw, do you guys know that they have Pokémon episodes on Netflix?
@@blackpenredpen Watch Deathnote BPRP.... You are too old for Pokemon 😂😂
@@050138, people who were 10 years old when Pokémon was released are now 34 years old. :)
Me: he's never going to finish it all on that little board...
blackpenredpen: *starts writing out even more intermediate steps*
This is the way I solved it:
At first I also assumed that u is equal to x^2, which gives you the following equation.
u^2-40u+q=0
If you then solve for u, you get this equation.
u=20+/-sqrt(400-q)
Now I used the fact, that x is the square root of u, which means, that the solutions for x aren't just a,b,c and d but -b, -a, a and b. Because the solutions have to form an A. P. b-a has to be equal to a-(-a):
b-a=a-(-a)
b-a=2a
b=3a
Now we know, that the four solutions for x are -3a, -a, a and 3a. The solutions for u have to be the squares of these four numbers, because I assumed, that x^2 is equal to u. If you square the four numbers you get a^2 and 9a^2. If we now come back to the formula for u, we know that 20+/-sqrt(400-q) is either a^2 or 9a^2 depending on whether you add or subtract sqrt(400-q). Because a^2 is positive you get 9a^2 if you add and a^2 if you subtract. Now you have two equations:
20+sqrt(400-q)=9a^2; 20-sqrt(400-q)=a^2
Now you can multiply the right equation by 9 to get 180-9sqrt(400-q)=9a^2. Then you can combine this equation with the left equation to get the following.
20+sqrt(400-q)=180-9sqrt(400-q)
Now you can finally solve for q:
sqrt(400-q)=160-9sqrt(400-q)
10sqrt(400-q)=160
sqrt(400-q)=16
400-q=256
144-q=0
144=q
Cool
Beautiful problem! Thanks for all you do, I am looking forward to watching new videos :D
what i did before watching:
let the roots be a, a+b, a+2b, a+3b
then * can be written as (x-a)(x-a-b)(x-a-2b)(x-a-3b)
by comparing coefficient of x^3, we get (-4a-6b)=0, b=-2a/3
so * can be re-written as (x-a)(x-a/3)(x+a/3)(x+a)
now compare the coefficient of x^2, we get a^2(-10/9) = -40, a=6
therefore, * is (x-6)(x-2)(x+2)(x+6)
q = (-6)(-2)(2)(6) = 144
Yes I did the same. The question is very straight forward and took 1min to solve. Then I saw he has solved in 10 min and thought my ans is wrong. But fortunately I am current.
Very brilliant solution
I solved it as follows:
The function is even so if r is the smallest positive root then -r is as well. Also 3r and -3r are roots (in order to form AP). It follows that:
Eq1: 81r^4 -360r^2 +q=0
Eq2: r^4 -40r^2 +q=0
(Eq1)-(Eq2) implies 80r^2 (r-2)(r+2)=0, Thus r=2 (positive)
So our original equation of the question becomes (x+6)(x+2)(x-2)(x-6)=0
Now it's clear that the constant term is 144 (which is q)
That was my solution too!
My first step was similar. But I calculated coefficient at x^2, knowing by Viet fotmula that it is a sum of all products of two roots. It equals -10*r^2 = -40, so r=2, so I know all roots and just multiplied them all by Viet again
Start with the sum. Since sum of roots is zero, the roots would have to be -3d, -d, d, 3d
Then, sum of pairwise products is -40, so -d^2 + -9*d^2 = -40
so d = +2, -2
Roots: -6, -2, 2, 6
q = (-1)^(degree of polynomial) * product of all roots. = (-1)^4 * (-6*-2*2*6) = 144
I like the solution here exploiting the symmetry, but this can also be done by brute force algebraic manipulation easily enough:
Solns a,a+b,a+2b,a+3b.
Product(x-a-kb)=x⁴-40x²+q=0.
Compare x³ terms: 4a+6b=0 => b = -⅔a.
Compare x² terms: 6a² + 18ab + 11b² = -40.
=> -10/9 a² = -40 => a = ±6, b = -±4.
Compare x⁰ terms: q = (-6).(-2).2.6 = 144.
Don't solve x, solve q
*Proceed with solving d first*
Your understanding of mathematics is really brilliant ❤️i would like to make my brain think like you💕
ok so let's try to solve it before i watch:
t = x^2
t^2 - 40t + q = 0
for 2 different solutions
(-40)^2 - 4*1*q > 0
1600 - 4q > 0
q < 400
-----------------------
for both to be positive from Viete's formulas
-(-40) / 1 > 0 (their sum)
and
q / 1 > 0 (their product)
So my guess is 0 < q < 400
I did what you did. Can you tell me why it would be wrong?
i also did this, i dont think its wrong (cos 144 is indeed between 0 and 400) but it just isnt complete because not all values of q between 0 and 400 give you arithmetic progression when you sub it into the original eqn
came for math, stayed for math
Easy way
a - 3d , a - d , a + d , a + 3d in ap with c.d = 2d
Sum of roots = 0
4a = 0
a = 0
Taking products 2 at a time
total combimations - 4 C 2 = 6
- 3d × - d + - 3d × d + - 3d×3d + - d×d - d×3d + 3d × d = - 40
- 10d^2 = - 40
d = +- 2
Nos. are - 6 , - 2 , 2, 6 or vice versa
I just made q=mn, and then made the equation into (x^2-m)(x^2-n)=0, then solved that to get +-sqrtm and +-sqrtn. Then I just did sqrtm-sqrtn = sqrtn-(-sqrtn). This gave m = 9n, plug that into m+n=40, get n=4, and m=36, so q=144. This solution took about 2 minutes to solve. Also, whether n > m or m > n doesn't affect the solution, as either way one of m or n will equal 9n or 9m.
I don't even use this kind of math in my classes anymore, I just watch the videos can you make them fun and I also love math, but mainly because of you.
If you start with a polynomial (x-a)(x-b)(x-c)(x-d) you have a+b+c+d=0
since it is an arithmetic progression, you can then prove that the roots have the form a, a/3, -a/3 and -a
so the polynomial becomes (x2-a2)(x2-a2/9)
and you get a2 = 36
and a4/9 = 144 = q
I found it easier to do Vieta on the original quartic with roots a1, a2, a3, a4 with a1=A, a2=A+r, a3=A+2r, a4=A+3r.
Then 4A+6r=0 (vieta rules, coefficient of x^3) => A=-3/2r
WLOG, a1 r=4 (since we assumed r>0, a
You could also use the sum of the roots of a quartic equation 1, 2 and 4 at a time to figure it out, which I found much easier
Alternative method:
Take the four solutions as a-3d, a-d, a+d, a+3d. Then sum of roots = 4a= -(coefficient of x^3/coefficient of x^4)= 0, which implies a=0. Then the product of roots taking two at a time, and summing it up is given by coefficient of x^2/coefficient of x^4, which leads us to the equation -10d^2 = -40, which gives d=±2. Then we have the solutions as -6, -2, 2, 6. Constant term/coefficient of x^4 = product of roots implies, q=-6*-2*6*2=144
a-3d, a-d, a+d, a+3d be 4 roots.
Step 1 compare coeff of x^3 with sum of roots
Result a=0
Step 2 compare coeff of x^2 with sum of product of roots taken 2 at a time
Result d=2
Step 3 compare coeff of x^0 (or constant term) with product of all roots
Result q=144
Do you know Michael Penn??? He's keep saying great like what you did on this vid. Anyway he's great in math as well. I followed hin here in yt just like you. You both are great. Thanks Sir Steve.
blackpenredpen: Don't solve for x
Also blackpenredpen: *solving for x^2*
Thank god Pikachu was there to approve it
Yea
More math olympiad video like this please!
Good one ,solution can also be done by sum of roots and sum of 2 roots at a time, and 3 roots at a time and product of roots also
There are many solutions for the equation itself, but only one q works for constant spacing when sorted; the rest have integers and radicals. Under 400, q=144 delivers all integer solutions not just sequential.
Thanks for inviting such an amazing Pikachu
Love your videos! Your a great inspiration!
Quarantine really do be taking away writing space from mathematicians :(
Take the 4 roots to be a-3d, a-d, a+d, a+3d. It makes it easier to solve.
Actually, once you find r1+r2 = 40, you know they must be the sum of two squares. It is quite easy to see by inspection that only 4 and 36 fit.
So the solutions must be +- 2 and +- 6. The sequence -6, -2, +2,+6 is arithmetic with d=4.
love u bro I try learning a lot from u. Keep it up and once again love u ❤️
In the thumbnail the coefficient of x^2 is -20 but in the video, it's -40.🤔
I actually solved it using the thumbnail, then watched the video...
Chigga snitching
Joshua Mark maybe there‘s a relation between 20 and 40... What were your solutions for q and the A.P.?
MathForLife I think he changed it
But then the solution wouldn't be an A.P. (-4,-2,+2,+4)
You continuously remind me what my teacher always say...*all questions are solvable*
I did it pretty much the same way. Towards the end looked different because I noticed that the square root of r2 equals 3 times the square root of r1, and so I substituted it into the r1+r2=40 formula directly, which gives you that 10r1=40, so r1=4, and r2=9r1 (or 41-r1)=36. I was effectively using the idea of “d” without it appearing on the paper.
I wish you hadn’t said the solution at the beginning. At first I had to train my brain not to think of factor pairs for 144.
The solutions could be generalised, If say the equation was x^4 - a x^2 + q = 0 and the AP difference of the solutions is d, then for all n, {d = 2n, a = 10n^2, q = 9n^4 }
What is the name of the formula at 4:35? I tried to google it but I must have spelled it wrong because I couldn't find it.
Google Vieta's formula it's very useful I recommend learning it
How do you write q?...
Thanks for all the calculus vids
If an easier solution that gives ALL the values for q :
Let u=x², we get u²-40u+q=0.
If we want 4 real solutions, we need 2 real solutions for u that will give 4 real solutions for x.
In order to get 2 real solutions for u, we need d = (-40)² - 4q > 0 *q < 400*
Therefore the answers are u = [40 +/- sqrt(d)] / 2.
As u=x², then, we get x = +/- sqrt(u). As u can take two real values, then x can take 4 real values if u>0 (the sqrt function is defined on [0,+infinity[).
This, means that : [40 +/- sqrt(d)] / 2 > 0 and after some calculations, we get *q > 0* .
Therefore, x^4 - 40x² + q = 0 has four real solutions when 0 < q < 400.
We can notice that q=144 is between 0 and 400.
XD
a youtuber thats good does not need any video editing and still can get 485k subscribers
Man this is so much beautiful 🥺 thank you ❤️
Simply take the roots as a-3d,a-d, a+d,a+3 d, the 4 roots in AP. Sum of the roots is zero, makes a=0 and the sum of their 6 products taken 2 at times equal to 40 solves the roots.
My solution:
First I wrote the roots as(solved the quadratic equation for x^2)
x1,2,3,4=+-sqrt(20+-sqrt(40^2-4q)/2)
Then I used Newton's method to solve the following system of nonlinear equations for (q,a,d):
x1=a or x1(q)-a=0
x2=a-d or x2(q)-a+d=0
x3=a-2d or x3(q)-a+2d=0
d is the common difference of the arithmetic progreasion.
(q, a, d) = (144, 6, 4)
love your q's
you should make a font based on your typography!
i just messed around on desmos for a bit and applied my knowledge of quadratic equations until i got the answer lol
Can u please do natural log and Euler's number videos please ...
My thought process:
1. Observe that the sum of four roots are zero because the coefficient of x^3 is zero. Let the roots be -3k, -k, k and 3k such that they form an ap with common difference 2k. Note that q = the products of roots = 9*k^4
2. Substitute y = x^2, the equation becomes a quadratic equation of y. By symmetry, the two roots of y corresponds to (3k)^2 = 9k^2 and k^2. As for each solution of y, there will be two values of x, differ by +/- sign.
3. The sum of root of the quadratic equation of y is 40. Therefore, 9*k^2 + k^2 = 10*k^2 = 40 ==> k = +/- 2. But the sign does not matter here as, again, because of symmetry.
4. q = 9*2^4 = 144 and the roots are -6, -2, 2 and 6. QED =]
I just factorize it to (x²-a)(x²-b). Then, by guessing, with a+b=40, I just put in 4 and 36 as they are square numbers. Then it turns out that it works
i just wrote a general solution that would form an arithmetic sequence in the form
(x^2-u^2)(x^2-(3u)^2) where u is the positive root with smallest magnitude (roots are -3u, -u, u, 3u, 2u between all of them)
expanded this out to get x^4-10u^2x^2+9u^4
the x^2 term is also -40 so 10u^2=40 => u=+ - 2 plus or minus doesn't really matter since its symmetrical so just gonna use +2
sub this into 9u^4 get 9x16=144
so q = 144 and the ending quartic is x^4 - 40x^2 + 144
its always nice seeing a solution straight away, especially on a bprp video
Just solved it for x:
x^4 - 40x^2 + q = 0
1.
let t = x^2, t>=0
t^2 - 40t + q = 0
t = 20 +- sqrt(400-q)
to have 4 different sol-s:
0 < q < 400
2.
numbers on the x axis are:
-sqrt(20+sqrt(400-q)), -sqrt(20-sqrt(400-q)), sqrt(20-sqrt(400-q)), sqrt(20+sqrt(400-q))
so if it's an AP then:
3 * 2 * sqrt(20-sqrt(400-q)) = 2 * sqrt(20+sqrt(400-q))
squaring:
9(20-sqrt(400-q)) = 20 + sqrt(400-q)
16 = sqrt(400-q)
squaring again:
256 = 400 - q
q = 144
profit.
0:42 quad...quartic
I see what you did there 😂😂
hey blackpenredpen, there's been a question ive been struggling on and i hope you can help me solve it. The question in full is:
15*(5/4)^(n-1) > 5n+10 and solve for the conditions for n. This question is related to the geometric and arithmetic nth terms, I skipped ahead and made the conditions for both sides. Please have a look at this for me
Brackpenredpen is good D+++++++++++++++++++♾
Can someone explain me (in a way that a common mortal can understand) how do we know that the difference between the sqrts of solutions is "d" and it is costant?
Me too, I can't understand that 🤔
It's just the definition
That hides in "the solutions follow an arithmetic progression", which means that the difference between two numbers, if written in ascending order, is always the same.
@@erynn9770 , Well, there is symmetry in the graph because of the arithmetic progression requirement, and symmetry is found in aesthetics.
Hi! I've been watching your videos for weeks and I am really fascinated. I love math and I really want to study it at advanced level and be good at it, but I don't know how to begin. Have you got any advices(maybe some videos or sites or books)?
I hope you read this message
Thank you, you're amazing
A site you could go to is Brilliant , they explain math in a very fascinating (I feel to say) way
What level of math education do you have?
@@JSSTyger I'm an italian student and I'm at the third year of high school(11th grade)
@@matniet43 I've just seen the site and I think it's very cool! I just have to check better for more information, but thank you
@@killergames7923 So i take it you have done basic geometry, trig, etc right? What about Calculus?
When I paused the video, I kinda stumbled across the solution by accident, before I started doing any real maths: I saw that the graph would be symmetrical, and would look like a W. So just to get an idea of the shape, I set q=0, and worked out the values from x=-3 to x=3. I noticed that -2 and +2 both gave -144. On a whim, I tried it for 6, and sure enough, -144. So then it was obvious that for q=144, you'd get solutions for x=0 at -6, -2, 2, and 6.
I was kinda disappointed - I feel like the Olympiad writers should have tweaked the equation a little, put different coefficients etc, to make the solution less easy to just stumble across.
I solved according to the thumbnail and found 18.
Sorry I just fixed my mistake
no problem
4:10 is Viette's relation
Why you holding that yellow whatever?
Nice solution! And very efficient use of the white board. :-)
My attempt ( first time attempting a blackpenredpen video question :D )
Let u = x^2 --> u^2 - 40u + q = 0
Using quadratic formula, solutions are (sqrt = square root)
--> u_1 = (40 + sqrt(1600 - 4q))/2 = 20 + sqrt(400 - q), and
--> u_2 = (40 - sqrt(1600 - 4q))/2 = 20 - sqrt(400 - q) .
The sqrts of these will be the four solutions. Since u_1 > u_2 > 0, they must be evenly spread around zero, and visualising this (i drew a number line), we see
--> sqrt(u_1) - sqrt(u_2) = 2 * sqrt(u_2)
--> sqrt(u_1) = 3 * sqrt(u_2)
--> u_1 = 9 * u_2
Subbing in the above expressions for u_1 and u_2:
--> 20 + sqrt(400 - q) = 9 * (20 - sqrt(400 - q))
--> 20 + sqrt(400 - q) = 180 - 9 * sqrt(400 - q)
--> 10 * sqrt(400 - q) = 160
--> 400 - q = 256
--> q = 144
:) #yay
You consider r1< r2 and got q=144
If you had consider r1=r2 you have got q=0
i delayed to realize the pikachu was actually the mic
Here's my go at it:
To get 4 real sols, the quadratic with z = x^2 needs two positive sols b > a > 0, so the 4 real sols are +/- sqrt(a) and +/- sqrt(b). Let r^2 = a and s^2 = b. We know the order is -s, -r, r, s, so the gap is r-(-r) = 2r and so s = r + 2r = 3r. That means the solutions to the quadratic are a and 9a. (z - a)(z - 9a) = z^2 - 10az + 9a^2 = z^2 - 40z + q, so a = 4 (and b = 36) and q = 144. Solutions are -6, -2, 2, 6.
BlackpenRedpen, I've watched your 'Integrating sec^n (x)' video. I actually have an interesting issue with it..
*Please look at this, people*
I derived the values of Integral of sec^2(X), sec^3(X) , ... till the 6th power of sec (X).
On the other hand, I verified the answers and differentiated them to check if they were right.
The issue is with the answer of Integral of sec^4 (X).
I derived the value of this problem with 2 different steps.
Int. sec^4 (X) = Int. sec^2(X) * sec^2 (X) dx.
I used the DI Method to work this out.
The answer will be,
sec^2(X).tan(X)
- 2 Int.(sec^2(X)*tan^2(X)*).
1st method:
I did 'U- Substitution' for tan(X) and ended up with,
sec^2 (X).tan(X)
- 2 Int.( u^2. du)
= sec^2(X).tan(X)
-2/3* tan^3(X)+ c.
When I differentiate this, I get sec^4(X), which is all perfect..
2nd method:
Instead of doing 'U-Substitution' for tan(X), I replaced tan^2(X) with
sec^2(X)-1.
And when I did the following, I ended up with the answer,
1/3*sec^2(X).tan(X) + 2/3*tan(X) + c
*which is also the answer derived by using the Int.sec^n(X) formula*..
When I differentiate this I end up with,
7/3*sec^2(X).tan^2(X), instead of sec^4(X)..
Please help me with this.
It took me really long to type this perfectly. It'll be worth it if I get a good response.
Had to comment in this video, because it's the Latest one.
Please help me. Thanks!
Can you explain how to use the Trachtenberg System?
(1) 1:36 . . .
Curious - Could complex numbers not form APs as well?
(2) 2:16 . . .
If all the roots are equal they also form an AP - a trivial case AP with zero common difference.
was waiting for a r2d2 joke :D
Dear all,
In this lockdown stage in home, please provide game equipments to your children/students to play. If not help them to watch "Math Art Studio" in you tube. They will play with their names and learn different concepts in mathematics.Those who have seen it they have learnt maths and enjoyed its beauty every day.
Thanks for the tips.
How about trying geometry and combinatorics problem from math olympiads
So if you assign x=2 and q=144 to the original expression it actually equals 0. But if you try with different values for x it doesn't work out anymore.
?
@@blackpenredpen I meant: x^4 -40x^2 + q = 0
Assigning the value 2 to x and the value 144 to q you actually make it work the result is 0. But if you assign a different value to x like for example 3 it doesn't work anymore, it doesn't give you 0 that expression.
So basically you solved for q but only for the case x=+-2 ?
Is anyone able to provide me an answer and let me understand where I may be wrong in this?
Make a vedio on SSC questions ..
P= 144
4 solns (+ - 2, + - 6)
Hey BPRP!! Can you solve this problem?
Find all primes numbers p and q such that p^3 - q^5 = (p + q)^2
Thank you so much!!
I found it also intesting to solve the problem with the four roots forming an GEOMETRIC progression...
Answer is then q = 800
I did in the form of AP from the start nd its coming 900. if we take the roots to be as a-3d , a-d , a+d, a+3d
thanks bro 👍👍🔥
@blackpenredpen plz some tough Indian Olympiad problems basically RMO ( Regional Mathematical Olympiad) and INMO ( Indian National Mathematical Olympiad)...
I would like them albeit you solve a few IIT JEE ADVANCED problems. Plz some as many integrals of IIT and some Indian Olympiad problems to help us Indians wo sometimes are unable to understand the solution.
Thank you
Regards
Abhijeet
Thanks for solutions. Salute from Turkey...👋
what is the slution?
Here is the Nike way :
x=a , a+c , a+2c , a+3c
then
a^4 - 40a^2=(a+c)^4 - 40(a+c)^2=(a+2c)^4 - 40(a+2c)^2=(a+3c)^4 - 40(a+3c)^2
after doing a little bit of algebra you will get 3 equations with coefficients :
4 6 4 1 -80 -40
8 24 32 16 -160 -160
12 54 108 81 -240 -360
make sure to put them in the same order then simplify to get
4 6 4 1 -80 -40
0 -12 -24 -14 0 80
0 0 24 36 0 0
the last one tells us that:
24ac^3+36c^4=0 --> 12c^3(2a+3c)=0
a=-3/2c since c cannot be 0
plug in the first equation to get
c=+-4 -->a=-+6 don't care which one is just plug any a into the original equation
6^4 - 40(6)^2 +q =0
q=144
first I made a mistake (forgot to square the 3 after plugging a=-3/2c) but then fixed it
honestly I tried for an hour didn't get it
XDDD
Por favor podría ayudarme con la integral definida de 0 hasta 1 de ((1-x^2)^1/2)ln(1-x/1+x)dx
I can't understand the middle 'd'.... please explain
If r1 and r2 are the roots of ** then why are the roots of * as root r1 and root r2.....please tell......btw i love ur videos
This was a complete mindfuck... but I loved it!
Why the spoiler at the beginning ?
Well, some ppl might have tried the problem already so I just wanted to tell them the numerical answer right away. On top of that, the numerical answer is never that interesting, but the solution is. : )
@@blackpenredpen I can't disagree with that
Toppers in imo before:😎😎😁😁
Blackpenredpen enters:
Le toppers:😱😱😱😅😅
"toppers"?
@@Daniel-nl3ug topper means the genius students or imo elites
❤
suppose that roots are
a - 3d , a - d , a+d , a+3d
since coefficient of x^3 = 0 , sum of roots = 0
or a = 0
now roots are
- 3d , - d , d , 3d
equation will be
(x^2 - 9d^2) (x^2 - d^2) = 0
or x^4 - 10 d^2 x^2+9d^4 = 0
comparing coefficient of x^2 and constant term
10d^2 = 40 , 9d^4 = q
d^2 = 4 hence q = 9× 16 = 144
what if i consider r1>r2 ?
is it gonna be the same?
Who is that blue pen guy?
I tried solving the qs in thumbnail for 10 mins, and saw -40x² in the video 😂😂😂
Same here
Indian🇮🇳👳 and Chinese🇨🇳👲 are always good at mathematics