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!
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...
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)
(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.
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.
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.
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 🥴
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|
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....!!!!!!
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
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
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!
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...
Another impressive video.
Thank you for another educational video
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)
(a+5)(b+5)(c+5)(d+5)=29
This question doesn't make sense
@@subramanyakarthik5843it does
(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.
Very good. Thanks Sir 👍
This was a great problem, do you do linear algebra
Sometimes. If it's not too complicated
That was very helpful!
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.
This was actually fun and I don't know why
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.
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 🥴
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|
That's the point
@@PrimeNewtons I rewatched and see it now. thanks
I think you should use quadratic equations to solve instead of AM-GM inequality as x was assumed to be negative
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
Aren’t (x,y,z)=(1,1,0) and (0,1,1) valid sets of solutions
No, second equation doesn't hold.
No. 1,1,0 doesn’t satisfy second equation
1^2-0≠0
Iam 16 and i try to solve it and finally i do it😆
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....!!!!!!
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
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
Great detail. Thank you 👍