This system of equations required showing that some equations were possible for real numbers and some where not.
Пікірлер: 33
@quzpolkasАй бұрын
You can show that x^2 + xz + z^2 = 0 has no real roots without using AM-GM inequality, by using properties of parabolas/quadratic polynomials. First, we show that x = z = 0 contradicts original equations 1 and 3, thus we can assume that x, z are not 0. Divide both sides by z^2, substitute x/z with a new variable t, we get: t^2 + t + 1 = 0 This quadratic polynomial's discriminant is D = 1^2 - 4*1*1 = -3 < 0. A quadratic polynomial with real coefficients has got real roots if and only if it's discriminant is non-negative, so this equation in t and, consequently, the previous equation in x and z don't have real roots. Cheers!
@pojuantsalo3475Ай бұрын
I thought I had this, but then I realized my solution doesn't make sense. I found the error and tried again, but I end up with a very nasty equation x^2 = 1 - (x^2 * (1-x^2)^3)^(1/4). At that point I gave up. These kind of problems require you take the exact correct first steps solving them or you are screwed. Of course this time it was to set equation (1) equal to equation (3). I feel stupid for not trying that, because the rest is pretty simple... What I tried was rewriting the equations as yz = 1-x^2 xz = y^2 xy = 1-z^2 And multiplying the all together: x^2*y^2*z^2 = (1-x^2)(1-z^2)y^2 = (1-x^2-z^2+x^2*z^2)y^2. If y ≠ 0, x^2*y^2 = 1-x^2-z^2+x^2*z^2. This got me into trouble in the end...
@jichehu1285Ай бұрын
notice that x and z can replace each other. use equation 1 and 2 we can get (x - y + z)(x - z) = 0. suppose x - y + z = 0, we can get x^2 + z^2 + xz = 0 and x^2 + z^2 + xz = 1, which is impossible. so x - z = 0 must be true. we can get x^2 = y^2 and x^2 + xy = 1, therefore 2x^2 + 2xy = x^2 + 2xy + y^2 = (x + y)^2 = 2 so x + y = 0 must be false, therefore x = y = z.
@scavengerethicАй бұрын
I got it by multiplying eq1 by x, eq2 by y, eq3 by z, then you have an xyz term in all of them you can subtract away and rearrange to prove x^3 - x = z^3 - z, which is only true if x=z. Then the middle equation means y^2 = x^2 = z^2 and so on from there.
@dougaugustine4075Ай бұрын
Another impressive video.
@JatinRohilla-mi6lmАй бұрын
Please solve bro newton if (a+1)(b+1)(c+1)(d+1) =1 and (a+2)(b+2)(c+2)(d+2)=2 and (a+3)(b+3)(c+3)(d+3)= 3 and (a+4)(b+4)(c+4)(d+4) =4 then find (a+5)(b+5)(c+5)(d+5)
@torlafkensom3093Ай бұрын
(a+5)(b+5)(c+5)(d+5)=29
@subramanyakarthik5843Ай бұрын
This question doesn't make sense
@ThoseInterestingStoriesАй бұрын
@@subramanyakarthik5843it does
@FirstNameLastName-mw1pjАй бұрын
(a+x)(b+x)(c+x)(d+x) =(ab+ax+bx+x^2)(cd+cx+dx+x^2) =abcd+(abc+abd+acd+bcd)x+(ab+ac+ad+bc+bd+cd)x^2+(a+b+c+d)x^3+x^4 Which is a quartic. You gave the value of the quartic at 4 points. We would normally require 5 points to find unique values for the coefficients for a quartic because it'd be a system of 5 equations and 5 unknowns, but because we know the x^4 coefficient should be 1, a system of 4 equations with 4 unknowns should give a unique soln. For the sake of me not having to type out that combination repeatedly, let's call (a+b+c+d)=s, (ab+ac+ad+bc+bd+cd)=t, (abc+abd+acd+bcd)=u, abcd=v. s+t+u+v=0 8s+4t+2u+v=-14 27s+9t+3u+v=-78 64s+16t+4u+v=-252 s+t+u+v=0 0s+4t+6u+7v=14 0s+18t+24u+26v=78 0s+48t+60u+63v=252 s+t+u+v=0 0s+4t+6u+7v=14 0s+9t+12u+13v=39 0s+16t+20u+21v=84 s+t+u+v=0 0s+4t+6u+7v=14 0s+t+0u-v=11 0s+0t+4u+7v=-28 s+0t+u+2v=-11 0s+0t+6u+11v=-30 0s+t+0u-v=11 0s+0t+4u+7v=-28 s+0t+u+2v=-11 0s+0t+2u+4v=-2 0s+t+0u-v=11 0s+0t+4u+7v=-28 s+0t+u+2v=-11 0s+0t+u+2v=-1 0s+t+0u-v=11 0s+0t+4u+7v=-28 s+0t+0u+0v=-10 0s+0t+u+2v=-1 0s+t+0u-v=11 0s+0t+0u+v=24 s+0t+0u+0v=-10 0s+0t+u+0v=-49 0s+t+0u+0v=35 0s+0t+0u+v=24 s=-10, t=35, u=-49, v=24 f(x)=x^4-10x^3+35x^2-49x+24 f(1)=1, f(2)=2, f(3)=3, f(4)=4, so it works for all points given. f(5)=625-1250+875-245+24=29 So (a+5)(b+5)(c+5)(d+5)=29, as torlafkensom said.
@Bitz00.Ай бұрын
Thank you for another educational video
@surendrakverma555Ай бұрын
Very good. Thanks Sir 👍
@evgeniospagkalis9922Ай бұрын
That was very helpful!
@BinduYadav-ydАй бұрын
I think you should use quadratic equations to solve instead of AM-GM inequality as x was assumed to be negative
@i18nGuyАй бұрын
AM-Gm is for non-negative numbers so I dont think it applies here. Consider if one of xz is negative then x**2 + z**2 can certainly be greater than -2|xz|
@PrimeNewtonsАй бұрын
That's the point
@i18nGuyАй бұрын
@@PrimeNewtons I rewatched and see it now. thanks
@bombergame8636Ай бұрын
This was actually fun and I don't know why
@lotgcАй бұрын
Oh, I guess I misunderstood what the question wanted. I found that if: X=0, 1, -1 Y=0 Z=1, -1 Then it satisfies the equations. I thought the question was just asking for all real answers. Although, it would still be wrong because I didn't think to look at non-whole numbers 🥴
@maxvangulik1988Ай бұрын
x^3+xyz=x y^3-xyz=0 z^3+xyz=z xyz=x-x^3=z-z^3=y^3 assume x=z y^2-xz=0 (given) y^2-x^2=0 y=+-x x^2+yz=1 (given x^2+-x^2=1 but x^2-x^2=0, so y=x 2x^2=1 x=+-sqrt(1/2) solution 1: +-(sqrt(1/2),sqrt(1/2),sqrt(1/2)) actually i think that's it
@XGames_MratedАй бұрын
This was a great problem, do you do linear algebra
@PrimeNewtonsАй бұрын
Sometimes. If it's not too complicated
@samuelreinsbergАй бұрын
Aren’t (x,y,z)=(1,1,0) and (0,1,1) valid sets of solutions
@quzpolkasАй бұрын
No, second equation doesn't hold.
@souverain1erАй бұрын
No. 1,1,0 doesn’t satisfy second equation
@maxvangulik1988Ай бұрын
1^2-0≠0
@user-ke6hz3sj9iАй бұрын
(-1,0,1)
@himadrikhanra7463Ай бұрын
Y^2 = XZ... X,Y ,Z are in gp Y/ x = Z /y X + yz / x= 1/x X + yz × z/ y^2 = z / y^2 X = z/ y^2 ( 1 - yz) = z / y^2 × x^2 X = y^2 /z 1) Y^2 = xz 2) Boundary condition? 6 variable 3 equation! Unable to progress I am....!!!!!!
@SidneiMVАй бұрын
y² = xz x² - z² - y(x - z) = 0 (x + z)(x - z) = y(x - z) x - z = 0 => x = z => y² = x² => y = ± x y = -x => rejected because x² - x² ≠ 1 => x = y = z 2x² = 1 => x = ± √2/2 *x = y = z = (√2)/2* or *x = y = z = -(√2)/2* x + z = y xz = y² t² - yt + y² = 0 (t are the roots x and z) t = (y/2)(1 ± i√3) => no real solutions another way x + z = y y² = xz y² = x² + z² + 2xz x² + z² = -y² => y = 0 x + z = 0 xz = 0 x = y = z = 0 => rejected because 0² + (0)0 ≠ 1
@problems_solverАй бұрын
Iam 16 and i try to solve it and finally i do it😆
@BRUBRUETNONOАй бұрын
Hi, Thanks for your interesting problem. Here is the way, I solved it ! Of course, I didn't look at your solution. Tell me if you like mine. Greetings. Recall of the system (i) x^2+yz=1 (ii) y^2-xz=0 (iii) z^2+xy=1 (i)-(iii) gives x^2-z^2-(x-z)y=0 then (x-z)(x+z)-(x-z)y=0 then (x-z)(x+z-y)=0 so we have two cases CASE A (x-z)=0 or CASE B (x+z-y)=0 LET SOLVE FOR CASE A x-z=0 then x=z. So we have (A) x=z (ii) y^2-z^2=0 => y^2=z^2 => z=+/-y In case A we have the two subcases, Case A1 x=z=y and Case A2 x=z=-y Case A1 solves as follows with (x;y;z)=(x;x;x) Let's look for x (i) x^2+x.x=1 (ii) x^2-x.x=0 => verified for all x (iii) x^2+x.x=1 => equivalent to (i) Then 2x^2=1 => x=+1/sqrt(2) or x=-1/sqrt(2) Then Case A1 has TWO SOLUTIONS BELOW (x;y;z)={[+1/sqrt(2);+1/sqrt(2);+1/sqrt(2)];[-1/sqrt(2);-1/sqrt(2);-1/sqrt(2)]} Case A2 solves as follows with (x;y;z)=(x;-x;x) Let's look for x (i) x^2-x.x=1 => 0=1 => which is illogical (ii) (-x)^2-x.x=0 => 0=0 +> verified for all x (iii) x^2+x.(-x)=1 => 0=1 => which is illogical Then Case A2 has NO SOLUTIONS LET SOLVE FOR CASE B x+z-y=0 then x=y-z. The system writes as below (B) x=y-z (i) (y-z)^2+yz=1 => y^2+z^2-yz=1 (iv) (ii) y^2-(y-z)z=0 => y^2+z^2-yz=0 => y^2-yz+z^2=0 (v) From (v) we can write (y-z/2)^2-z^2/4+z^2=0 => (y-z/2)^2+3/4.z^2=0 => As we look for real values, the sum of the two squares (y-z/2)^2 and 3/4.z^2 being equal to zero, then these two values have to be equal to zero each => y-z/2=0 and z=0 then y=0 => from (B) x=y-z => x=0 Then x=y=z=0 which doesn't fulfill the equation (i) x^2+yz=1 and (iii) z^2+xy=1 of the initial system. Then CASE B has NO SOLUTIONS. To conclude, the following system (i) x^2+yz=1 (ii) y^2-xz=0 (iii) z^2+xy=1 has the following solutions (x;y;z)={[+1/sqrt(2);+1/sqrt(2);+1/sqrt(2)];[-1/sqrt(2);-1/sqrt(2);-1/sqrt(2)]} END