This is a really nice problem that uses symmetry in its solution! This strategy can be applied to very many different situations. It is not hard compared to some of the other algebra problems we tackled. Any thoughts on the problem or the solution method?
I like how noticing the opportunity for symmetry simplified things. Another way would have been to use guess-and-check, use the guessed solutions to help factor the quartic, and see that there are no other real solutions. But then we would not have had the lesson in symmetry.
First of all, put the LHS over a common denominator: [(x+2)² + x²] / x²(x+2)² = 10/9 => (2x² + 4x + 4) / x²(x+2)² = 10/9 There is a fraction on both sides of the equation, we can thus let the numerators and denominators be equal to eachother and compose the following system of equations: (1) 2x² + 4x + 4 = 10 (2) x²(x+2)² = 9 We can divide (1) by 2 and take the square root of (2), this then becomes the following: x² + 2x - 3 = 0 x² + 2x - 3 = 0 I.e. both conditions are equivalent. We can solve this quadratic equation and conclude that the only x satistying this condition are {-3, 1}.
Nice video!! I did the same substitution that you did, but I was not thinking about average, but the nice product (u - 1)(u + 1). And instead of turning the equation into a biquadratic I completed the square and transformed the equation into y^2 + y = 10/9 Where y = 2/(u^2 - 1) And proceed from that. I learned this kind of substitutions in other of your videos :). Thank you for sharing this content!!
The solutions x = 1 and x = -3 can be found from inspection. Consideration of the graph shows no possible solution between the two asymptotes and just the two solutions already found. Algebra can be used to confirm no further real solutions.
good evening It is also possible to do the sum and equal numerator with denominator with what we would have x^2=10 and (x+2)^2=9 if we do y=x^2 and z=(x+2)^2 we get the solutions y=9 or y= 1 as well as z. From here the solutions for x are obtained, leading us to x=1 or x=-3
The rhs is clearly 1/1 + 1/3^2 So i guessed x=1 as a solution. I didn't see x=-3 , but i should have done especially as equations involving squares typically have two solutions
hello dear teacher : i tried another way , hoping is a good way ! I let (1/x) = a & 1 / (x+1) = b .. So i hv a system with: (a^2 + b^2) = 10/9 and (b - a) /ab = 2. I know that: (a^2 + b^2) = (a - b)^2 + 2ab (b -a) = - (a - b) = - 2ab and obtain from that : 9 (a^2 + b^2) = 10 --> 9 [ (-2ab)^2 +2ab ] =10. Calling ab = v I resolve and obtain : 36 V^2+ 18 V -10 = 0 --> ab = - 5/ 6 or V = 1/ 3. by these I find a = 1/ 2 and b = - 7 / 6
before seing your solution i Just wrote 10/9 = 1 + 1/9. Next i tried by analogy of the two members of the equation : 1/x**2 = 1 which gave me x =1 and 1/(x+2)**2 = 1/9 . Next looking for another solution in a symetrical manner, I tried 1/(x+2)**2 =1 which led me to x = (-3) and : 1/x**2 = 1/9 . of course in 3mn mentally, I could not prouve that they were the only solutions. Thank you for your videos that teach me a lot!
i did it again and as the second member is 1+ 1/9 , either 1/x^2 is 1 and 1/(x+2)^2 is 1/9 or the inverse and to have (x+2) = 1 we need x= -3 the second solution. thanks a lot, i do have fun with your problems.
When you work out, you don't just bench press a car. You do less impressive things, like run and squeeze and other nonsense. You've got to tackle hard problems for strength/endurance, but you also need to solve easier problems for form and core strength.
Also hey SyberMath ! Could you make some more videos on questions based on Greatest Integer Function , Ceiling Function & Fractional Part Function ? I mean questions which have all these 3 functions mixed in them !
Also how much do you think you can get in this exam . Check out the paper then tell me ! www.mtai.org.in/wp-content/uploads/2021/01/IOQM-question-paper.pdf
This poor little thing is so simple but you completely overcomplicated the problem. So my solution (of course way faster): Let's look at the RHS: 10/9=1+1/9... wait a minute... this is equal to 1/1+1/9=1/1²+1/3²=1/(-1)²+1/(-3)² So we just found two solutions: -3 and 1. Now let's introduce y=x+1 and let's define f(y)=1/(y-1)²+1/(y+1)² Our equation becomes: f(y)=10/9. We notice that f is an even function so we only need to study f on [0;1+infinite[ Between 0 and 1, we have: y-10 then the minimum value of 1/(y+1)² is obtained for y=1 and this value is 1/4. So f(y)>1+1/4>10/9 for any value between 0 and 1. For y>1, f is the sum of two functions that strictly decrease and are continuous, with an infinite limit for y=1 and a limit equal to 0 for y--->+infinite. Then any positive value is reached by f (theorem of intermediate value), and this value is obtained only once (f strictly decreasing). So the solution we found x=1 or y=2 is unique for positive values of y. And since f is even the solution we found x=-3 or y=-2 is unique for negative values of y.
This is a really nice problem that uses symmetry in its solution! This strategy can be applied to very many different situations. It is not hard compared to some of the other algebra problems we tackled. Any thoughts on the problem or the solution method?
Yes, this is really nice problem.
There is another method.
1/x^2+1/(x+2)^2=10/9
Put x+1=t
1/(t-1)^2+1/(t+1)^2=10k^2/9k^2
(2t^2+2)/(t^2-1)^2=10k^2/9k^2
2t^2+2=10k^2
t^2+1=5k^2
t^2=5k^2-1
(t^2-1)^2=9k^2
t^2-1=3k,. t^2-1=-3k
t^2=3k+1
5k^2-1=3k+1
5k^2-3k-2=0
5k^2-5k+2k-2=0
(5k+2)(k-1)=0
k=1
t^2=3k+1
t^2=4
t={-2,+2)
t=x+1
x={-3,1}
I like how noticing the opportunity for symmetry simplified things.
Another way would have been to use guess-and-check, use the guessed solutions to help factor the quartic, and see that there are no other real solutions. But then we would not have had the lesson in symmetry.
Symmetry is awesome!
First of all, put the LHS over a common denominator:
[(x+2)² + x²] / x²(x+2)² = 10/9
=> (2x² + 4x + 4) / x²(x+2)² = 10/9
There is a fraction on both sides of the equation, we can thus let the numerators and denominators be equal to eachother and compose the following system of equations:
(1) 2x² + 4x + 4 = 10
(2) x²(x+2)² = 9
We can divide (1) by 2 and take the square root of (2), this then becomes the following:
x² + 2x - 3 = 0
x² + 2x - 3 = 0
I.e. both conditions are equivalent.
We can solve this quadratic equation and conclude that the only x satistying this condition are {-3, 1}.
The complex solution are (i/√5)-1 and -(i/√5)-1.
right?
right !
Nice video!!
I did the same substitution that you did, but I was not thinking about average, but the nice product (u - 1)(u + 1).
And instead of turning the equation into a biquadratic I completed the square and transformed the equation into y^2 + y = 10/9
Where y = 2/(u^2 - 1)
And proceed from that.
I learned this kind of substitutions in other of your videos :).
Thank you for sharing this content!!
Nice! Thanks for sharing!
Isn't x = 1 obvious from the initial equation after cross-multiplying though?
What book do You recommend for find this type of problems??
Good question! Hard to find a single book that contain problems like these but I can recommend some problem solving books.
@@SyberMath yeah!! Excellent!!. I Will hope You recomendations ♥️♥️♥️
@@SyberMath please share here or on Twitter. Following you here and there
Sir can you pls recommended your problem solving books@@SyberMath
The solutions x = 1 and x = -3 can be found from inspection. Consideration of the graph shows no possible solution between the two asymptotes and just the two solutions already found. Algebra can be used to confirm no further real solutions.
Beautiful solution. symmetry is such a powerful tool in maths at all levels of physics.
Thank you!
Which application are you using to solve this maths problem
Notability
good evening
It is also possible to do the sum and equal numerator with denominator with what we would have
x^2=10 and (x+2)^2=9
if we do y=x^2 and z=(x+2)^2 we get the solutions y=9 or y= 1 as well as z. From here the solutions for x are obtained, leading us to x=1 or x=-3
Nice!
how do you write it? is it like a chalk board? in which software u use to record?
I use screen recording on iPad with the Apple Pencil. I write in Notability.
The rhs is clearly
1/1 + 1/3^2
So i guessed x=1 as a solution. I didn't see x=-3 , but i should have done especially as equations involving squares typically have two solutions
That's right!
hello dear teacher : i tried another way , hoping is a good way !
I let (1/x) = a & 1 / (x+1) = b .. So i hv a system with: (a^2 + b^2) = 10/9 and (b - a) /ab = 2. I know that:
(a^2 + b^2) = (a - b)^2 + 2ab
(b -a) = - (a - b) = - 2ab
and obtain from that : 9 (a^2 + b^2) = 10 --> 9 [ (-2ab)^2 +2ab ] =10. Calling ab = v I resolve and obtain :
36 V^2+ 18 V -10 = 0 --> ab = - 5/ 6 or V = 1/ 3. by these I find a = 1/ 2 and b = - 7 / 6
10/9 = 1 + 1/9 or 1/9 + 1
Equating the fractions gives us both values of x = 1, -3
before seing your solution i Just wrote 10/9 = 1 + 1/9. Next i tried by analogy of the two members of the equation : 1/x**2 = 1 which gave me x =1 and 1/(x+2)**2 = 1/9 .
Next looking for another solution in a symetrical manner, I tried 1/(x+2)**2 =1 which led me to x = (-3) and : 1/x**2 = 1/9 .
of course in 3mn mentally, I could not prouve that they were the only solutions.
Thank you for your videos that teach me a lot!
i did it again and as the second member is 1+ 1/9 , either 1/x^2 is 1 and 1/(x+2)^2 is 1/9 or the inverse and to have (x+2) = 1 we need x= -3 the second solution.
thanks a lot, i do have fun with your problems.
This question was relatively easier than others 😁
Yes. Sometimes it's good to have an easier problem!
When you work out, you don't just bench press a car. You do less impressive things, like run and squeeze and other nonsense. You've got to tackle hard problems for strength/endurance, but you also need to solve easier problems for form and core strength.
@@Qermaq Thanks for the insight !
@@Qermaq Good approach!
@@SyberMath
Also hey SyberMath ! Could you make some more videos on questions based on Greatest Integer Function , Ceiling Function & Fractional Part Function ? I mean questions which have all these 3 functions mixed in them !
Sure. Why not?
Ty for your nice solution
Hello Lexi :)
@@Icewallocumm Hi
@@Icewallocumm Hi
This one is so easy. Look at rhs = 10/9
Denominator =9 = 3^2
Assume( x+2)^2 = 9
X+2 = 3
X=1
1/1^2 + 1/ 3^2 = 1 + 1/9 = 9+1/9 = 10/9
nice solution thanks
No problem! Thanks!
You can also complete the square and then take 2/x(x+2) = u.
How?
@@SyberMath Just add and subtract 2/x(x+2) and complete (a-b)^2. Term in bracket becomes [2/x(x+2)]^2.
graceful
Great video
Thank you!
Also how much do you think you can get in this exam . Check out the paper then tell me !
www.mtai.org.in/wp-content/uploads/2021/01/IOQM-question-paper.pdf
Take 1/x =a, 1/x+2 = b, get a quadratic equation in t=ab. It will be more easier..
I use another méthode a^2-b^2 with 1/9 and 1 so easy
How?
Elegance!!!
💖
1/(x)=a;1/(x+2)=b and the rest is easy to solve... many problems solved in this fashion only
x= -1+i/√5 and x= -1-i/√5 also solutions
(y-4)(5y+1)=0 This is obvious.
Why?
Because the y equation is factorable so we don't need to use the formula.
👍👍👍👍👍
😊😊😊
you make substitution method easy... hehehe
😁
This poor little thing is so simple but you completely overcomplicated the problem. So my solution (of course way faster):
Let's look at the RHS: 10/9=1+1/9... wait a minute... this is equal to 1/1+1/9=1/1²+1/3²=1/(-1)²+1/(-3)²
So we just found two solutions: -3 and 1.
Now let's introduce y=x+1 and let's define f(y)=1/(y-1)²+1/(y+1)²
Our equation becomes: f(y)=10/9.
We notice that f is an even function so we only need to study f on [0;1+infinite[
Between 0 and 1, we have: y-10 then the minimum value of 1/(y+1)² is obtained for y=1 and this value is 1/4.
So f(y)>1+1/4>10/9 for any value between 0 and 1.
For y>1, f is the sum of two functions that strictly decrease and are continuous, with an infinite limit for y=1 and a limit equal to 0 for y--->+infinite.
Then any positive value is reached by f (theorem of intermediate value), and this value is obtained only once (f strictly decreasing).
So the solution we found x=1 or y=2 is unique for positive values of y.
And since f is even the solution we found x=-3 or y=-2 is unique for negative values of y.
10/9 =1+1/9
so x=1 or -3
i can solve by intuition at 3 seconds. haha
You are a genius! 😁
So did I as well
You can take : x+1=t
That's nice!
courtic
Whoelse try to "scroll up" the video?
Wow
🌝⭐🌟⭐
😊🥰😊