Different approach : You can multiply (a+b) on both sides of second equation ( a^2 + b^2 )=7 you will get a^3 + b^3 + (a^2)b + (b^2)a = 7(a+b) send a^3 + b^3 to RHS, take ab common in LHS, substitute (a+b)=x and a^3 + b^3 as 10 you get ab(a+b) = 7x-10 we know (a+b)^3 = a^3 + b^3 + 3ab (a+b) From previous eq substitue ab(a+b) x^3 = 10 + 3 (7x-10) you got the equation!!!
The answer was not what I expected. So I don't know what finding `a+b=?` really means. I thought that once we found that, any (a,b) that satisfies would also satisfy the original equations. This is clearly not the case. So do I interpret this as any solution to the system equations will also satisfy a+b=1, and that we cannot "elevate" a+b=1 to be part of the system?
Any (a,b) found satisfies the original equations, so you're mistaken that that is not the case. However, only the solution pair (a,b) for which a + b = 1, ab = −3 satisfies the requirement that a and b must be real, so this is the only solution to the problem.
I got three answers for a+b ,i.e., 1,-5 and lastly 4. Here's how I solved it, a²+b² = 7 => (a+b)²-2ab = 7 a³+b³ = 10 => (a+b)³-3ab(a+b) = 10 Now, let (a+b) and ab equal to x and y respectively. We get, x²-2y = 7 ---- eq. (1) x³-3xy = 10 ---- eq. (2) Upon solving this system off equations, we get (x,y) = (1,-3) , (-5,9) and (4,4.5) So, we get (a+b) = 1 or -5 or 4 Correct me, if I am wrong.
Great upload, Beautifully presented your precious video, I really appreciate your dedicated mind to create awesome contents for us, very well done, keep going, have a fantastic day to you and your family, lk447
I did it this way.let (a+b) =x, ab=y then we get x^3--3xy=10 (A)and x^2--2y=7(B) now multiplying equnB by x we get x^3--2xy=7x.now(A) --(B) is--xy=10--7xor xy=7x--10.putting this in equnA we get x^3--3(7x--10) =10 or x^3--21x+20=0 or (x-1) (x^2+x--20) =0 or(x--1) (x+5) (x--4) =0 but only (x--1) =0 satisfies or x=1 (,y=--3)i.e a+b=1.
Yes, another way is to multiply (A) by 2 and (B) by 3x and subtract to get rid rid of xy part. Then we get directly x^3--21x+20=0 and the result accordingly. Thanks.
Lengthy but was able to get it. Also, midway of the video, for x³-21x+20=0, instead of cubic grouping, I used a combination of 3 methods; 1.) Rational Zero Theorem 2.) Factor Theorem and 3.) Synthetic Division It may take some time due to finding possible zeroes, but I was able to find it. I also used a combination of these 3 methods when graphing polynomial functions. Back to the video, the rest is easy. And also, after finding the 3 quadratic equations, I paused the video and solved it on my own.
Great job, Kevin. Great approach on the Synthetic Division. Thank you for sharing! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!
Nice, very nice system of exponential equations. Thank you for the interesting math lesson and even more for the spoken English one, sir! God bless and I wish you to be always happy! Greetings from Ukraine:))
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I love how you select questions and solve them with clarity. One question one video.....awesome
That was comparable to watching a fine symphony play out. Well done!
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Different approach :
You can multiply (a+b) on both sides of second equation ( a^2 + b^2 )=7 you will get a^3 + b^3 + (a^2)b + (b^2)a = 7(a+b)
send a^3 + b^3 to RHS, take ab common in LHS, substitute (a+b)=x and a^3 + b^3 as 10
you get ab(a+b) = 7x-10
we know (a+b)^3 = a^3 + b^3 + 3ab (a+b)
From previous eq substitue ab(a+b)
x^3 = 10 + 3 (7x-10)
you got the equation!!!
What a wonderful question, how unusual it is, as if it came from another world...
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14:15
After validating the solution for a^2+b^2=7
Why you don't need to validate the solution for a^3+b^3=10 ?
it'll automatically be valid if it's in the first case.
Used a different approach. nice problem
a+b=1 answer
Since a^3+ b^3 = (a+b)(a^2+b^2-ab)
then (a+b) = 10/(7-ab) given that a^2+b^2=7 and a^3+b^3=10
since (a+b)^2 = a^2+b^2 + 2ab then
a+b = sqrt ( 7 + 2ab)
hence
sqrt (7+2ab) = 10/(7-ab)
7+2ab = 100/ ( a^2b^2-14ab +49) square both sides
let n=ab hence
7n^2-98n+343+2n^3-28n^2+98n =100
2n^3 -21n^2+243 =0
n= -3 satisfy the above equation -54 -189 + 243=0
or it can be factored into -27n^2 + 6n^2 + 2n^3 + 243=0
-27n^2 +243 + 2n^3+6n^2 since -27n^2 + 6n^2 = -21n^2
-27(n^2-9) +2n^2(3+n)=0
-27 (n+3)(n-3) +2n^2(n+3)=0
(n+3)[-27(n-3) + 2n^2]=0
hence n=-3
hence ab= -3
hence
a+b = 10/(7-(-3)
a+b = 10/10
a+b = 1 answer
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sir Your explanation skill is another level👊
fully understand your step-by-step solving this system of equations bro, thanks so much for sharing this problem
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Excellent... Pleasant to watch 🙌🏾
What would (a^4)+(b^4) be, if it’s possible to calculate? Given that a+b=1, (a^2)+(b^2)=7, and (a^3)+(b^3)=10?
What about actual possible values of a and b? What specific values of a and b can result in a*a + b*b = 7, and a*a*a + b*b*b = 10?
(1±√13)/2
You really know your stuff. You did a great job. 242 Math sent me.
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you are also right but can we use the elimination method then we get A+B = 3 easily?
Thanks
Amazing and brain-tickling problem. Thanks, Sir.
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The answer was not what I expected. So I don't know what finding `a+b=?` really means. I thought that once we found that, any (a,b) that satisfies would also satisfy the original equations. This is clearly not the case.
So do I interpret this as any solution to the system equations will also satisfy a+b=1, and that we cannot "elevate" a+b=1 to be part of the system?
Any (a,b) found satisfies the original equations, so you're mistaken that that is not the case. However, only the solution pair (a,b) for which a + b = 1, ab = −3 satisfies the requirement that a and b must be real, so this is the only solution to the problem.
i think, the exact question is: if a, b are real numbers and (!!) if a and b solve the two equations what is then x= a +b ?....
Absorbing indeed! - brings out curiosity
Thanks for video. Good luck sir!!!!!!!
Thank you too.
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I got three answers for a+b ,i.e., 1,-5 and lastly 4.
Here's how I solved it,
a²+b² = 7 => (a+b)²-2ab = 7
a³+b³ = 10 => (a+b)³-3ab(a+b) = 10
Now, let (a+b) and ab equal to x and y respectively.
We get,
x²-2y = 7 ---- eq. (1)
x³-3xy = 10 ---- eq. (2)
Upon solving this system off equations, we get (x,y) = (1,-3) , (-5,9) and (4,4.5)
So, we get (a+b) = 1 or -5 or 4
Correct me, if I am wrong.
Very interesting math question👍
Thank you so much for sharing😊
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wow incredible
Jaw-droppingly brilliant!
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Vvv nice stepwise explanation
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Great upload, Beautifully presented your precious video, I really appreciate your dedicated mind to create awesome contents for us, very well done, keep going, have a fantastic day to you and your family, lk447
Many many thanks
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Very good. Many thanks
Lovely solution
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Another Solution:
Because x, y ∈R,
(x-y)^2 ≧ 0 ⇔ (x+y)^2 ≧ 4xy.
Others are the same method except using the discriminants.
I did it this way.let (a+b) =x, ab=y then we get x^3--3xy=10 (A)and x^2--2y=7(B) now multiplying equnB by x we get x^3--2xy=7x.now(A) --(B) is--xy=10--7xor xy=7x--10.putting this in equnA we get x^3--3(7x--10) =10 or x^3--21x+20=0 or (x-1) (x^2+x--20) =0 or(x--1) (x+5) (x--4) =0 but only (x--1) =0 satisfies or x=1 (,y=--3)i.e a+b=1.
Many approaches possible!
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Yes, another way is to multiply (A) by 2 and (B) by 3x and subtract to get rid rid of xy part. Then we get directly x^3--21x+20=0 and the result accordingly. Thanks.
Easier if u start with (a + b ) ^ 3 .
It ends up : ( a + b ) ^ 3 = 21 ( a + b ) + 20
Then just a + b = x 😉
Since a and b are real numbers, we have
0
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Let's say we have 3 questions about that question, and the total questions are 50,they give us 100 minutes, how do we suppose to solve that faster?
Thanks sir
great problem
Thnku
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Lengthy but was able to get it. Also, midway of the video, for x³-21x+20=0, instead of cubic grouping, I used a combination of 3 methods;
1.) Rational Zero Theorem
2.) Factor Theorem and
3.) Synthetic Division
It may take some time due to finding possible zeroes, but I was able to find it.
I also used a combination of these 3 methods when graphing polynomial functions.
Back to the video, the rest is easy. And also, after finding the 3 quadratic equations, I paused the video and solved it on my own.
Great job, Kevin.
Great approach on the Synthetic Division.
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You are awesome. Keep it up 😀
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True, I did the same, but got the ans,
Nice, very nice system of exponential equations. Thank you for the interesting math lesson and even more for the spoken English one, sir! God bless and I wish you to be always happy! Greetings from Ukraine:))
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Greetings to you from the UK
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pretty good!!
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I confused about a = ? b = ?
Can somone explain me why I can not divede a^3+b^3=10 with
A^2+b^2=7
As than a+b=10/7?
you can divide: but a^3 +b^3 is not (a+b) * (a^,2 + b^2)...
you must know the rules of calculation....
I guessed 1 a few seconds after looking at the problem then did the work and got 4. I shoulda just went with my guess.
Почему Вы никогда не используете теорему Виетта?
It is an a cube equation so a and b should have 3 solutions. But your answer you got only two answers for a and b. Why?
Странно... Я получила три ответа: 1, -5 ; 4. И ни один не подходит. Числа a , b не принадлежат R. Нет решений?
Ошиблась в знаке при проверке. Подходит 1!
Без проблем. Мы все люди. Так держать.
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but chatgpt says a+b = 3
Complicate, but interesting equation
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a+b=1
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x=-5
Ans a = 1 , b = -3
really?
-3,5533482
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