this is a middle school level question in my country, I am now 20 and I learned this when I was 16, still remembering the best approach to solve this question just shows how goated my teacher is.... Love you Sir Navin Kumar
With the numbers offered this is a simple mental arithmetic problem. No pen or paper needed. 7 = (3 + 4) or (2 + 5) or (1 + 6) I went for 3 + 4. For the first square root to be 3 then X = 14 (14 - 5 = 9 and the square. root of 9 = 3) For the second square root to be 4 then X = 14 (14 + 2 = 16 and the square root of 16 = 4) Therefore X = 14 I did not bother checking the other two to see if they had a solution.
@@afsoc4life when they set these types of questions you would hope they would make the problem meaningful in terms of the numbers they use that would justify the need for complexity.
I'm with you in this. X=14 by inspection. No pen or paper, 15 seconds but faster for people who think faster than me, which isn't hard. Same route, what adds to 7, can you make it happen, yes? Done. The only problem with 15 seconds is if you try 1+6 or 2+5 first, and then confirming they don't work and moving to 3+4 would easily push me past 15 seconds.
The steps are written to those who are not at a proficiency level, maybe a beginner or intermediate. Certainly, a proficient person like you, would think these steps are unnecessary.
I think so too. When we solve such an equation, we use the concept called 自明解, which means evident solution. In this equation, we can clearly see one of the solutions is 14(this is 自明解), and then what we have to do is to proof that this is the only solution. This is easy because it is clear that when X increases, √X steadily increases. I don't know what this is called in English, but we call this 単調増加. 単調 means monotone and 増加 means increasing
Giải kiểu nầy sẽ bị trừ điểm vì thiếu điều kiện để phân thức trong căn tồn tại , nên ta phải đặt điều kiện ( Đk ) để phân thức trong căn tồn tại . Đk : x-5>=0 X+2>=0 Vậy : x >= 5 ...........
I just asked myself, "How do you get 7 from addition?" 7+0, 6+1, 5+2, and 4+3. And which are the closest together? 4+3. Then I squared both numbers to get 16 and 9. Then it was just simple math to get that 14-5=9 and 14+2=16, meaning x=14.
Sir, I dont get it First you choose 4 and 3 In my head based on your choice X - 5 = 16 , making x = 21 X + 2 = 9 , making x = 7 And is not the solution Or everthing is based on luck ?
@erwinyonathan6348 Because you need to swap it. x-5=9, x+2=16. It only works if you subtract 5 to get the lower number and add 2 to get the higher one. You'll never get x if you subtract 5 to get the higher number and add 2 to get the lower one.
Your process of solving the equation matched with mine line by line. And i don't think this problem to be too complicated. It's too easy for a 6th grader.
I have realized there are 3 main approaches 1. (a) Selecting the term / terms to be squared (b) Making the adjustments for the term / terms to be squared (c) Simplification 2. Use inspection to eliminate irrational solutions on the LHS : prior comment shows only 1 rational solution exists, the roots of the perfect squares ( 9, 16). 3. Reconstruct the method of problem construction : Abacus users have sight comparison tricks here. (a) The LHS is the sum of 2 integers, and their sum is the RHS integer. (b) The LHS is sum of the roots of 2 perfect squares. (c) The difference between the 2 perfect squares is the difference of the 2 constants = 7. (d) Investigate the perfect squares from 1 upwards value square 1 1 2 4 3 9 4 16 5 25 The pair of perfect squares with a difference of 7 is (9, 16). The sum of roots of the perfect squares is 3 + 4 = 7 (Value of RHS) For the smaller term : root( x - 5), x - 5 = smaller perfect square (9). x = 14. VERIFY For the larger term : root( x + 2), root (14 + 2 = 16) = 4. CONCLUSION X = 14 NOTE 1. I don't doubt someone will say I have been very wordy, but my aim was clarity over brevity. 2. There was something about the problem which made me think an abacus user would have a huge advantage , even if they claimed they didn't. 3. The reconstructing the construction method should work for any integer power, but for odd integer powers, the negative values demand evaluation as well.
@@MasterTMathsClass Thank you for the appreciation. I realize now why I suspected an abacus was used in construction. The sum 7, the difference of the constants 7, sum to the answer 14. Very like the symmetry in a Japanese stone garden.
Chasing perfect squares in each radical, you quickly find x=14 as an obvious solution. The function being increasing on its domain, the solution is unique. You could also take the square on the left and right, move the non sqrt terms on one side, square again, develop and solve the quadratic. Lots of work for nothin'
I suspect the Japanese are playing a trick on everyone. The result, RHS, is an integer 7. The LHS is the sum of 2 roots. If the roots are irrational, the sum will be also irrational. The possible integer sums adding up to 7 are (a) 1,6 (b) 2,5 (c) 3, 4 (d) 4, 3 (e) 5, 2 (f) 6, 1 Considering only the 1st term of the LHS, the possible values of X are (a) root(X - 5) = 1, squared = 1, giving X = 6. (b) root(X - 5) = 2, squared = 4, giving X = 9. (c) root(X - 5) = 3, squared = 9, giving X = 14. (d) root(X - 5) =4, squared = 16, giving X = 21. (e) root(X - 5) =5, squared =25, giving X = 30. (f) root(X - 5) = 6, squared =36, giving X = 41. Considering the 2nd term of the RHS, the valid value of X is (a) root(6 + 2) which is root(8) = 2 root(2) which is irrational (b) root(9 +2) which is root(11) which is irrational (c) root(14+2) which is root(16) = 4, an integer value. (d) root(21+2) which is root(23) which is irrational (e) root(30+2) which is root(32) = 4 root(2) which is irrational (f) root(41+2) which is root(43) which is irrational Only X =14 has integer roots, 3+4 = 7, the solution is X = 14. With the help of an abacus the possibilities can be explored in seconds. I feel sure the solution I have outlined is a clumsy version of the numerical tricks an abacus user has available.
by guesswork x=14 is a solution. on the lhs we have a sum of two monotonically increasing functions. lie the lhs is monotonically increasing function. Thus, the is only one value of x equal to any given x thus x=5 is the only real solution.
Acá independiente de la manera que se resuelva, algunas mas cortas otras mas larga lo importante es usar la creatividad, yo lo hice de una manera mas corta usando variables auxiliares, de todas formas hay que agradecer lo que se expone en el vídeo, dejar los eggo a un lado, porque todo sirve incluyendo los comentarios, que también aportan al desarrollo
You made your solution too complicated and unnecessary. The problem is very simple. There are several such methods. One is simply bu transposing, you can write Sqrt(x-%) = 7- Sqrt(x+2). Now squaring both sides and simplyfying, you get x+2 = 16; which gives your answer: x = 14. Dr. Ajit Thakur (USA).
rt(x+2)=7-rt(x-5)
x+2=49-14rt(x-5)+x-5
14rt(x-5)=42
rt(x-5)=3
x-5=9
x=14
You cooked 😅
Def: [1] A²=x+2
Def: [2] B²=x-5
=> [3] A+B=7 (A,B≥0)
[1]-[2]= 7=A²-B² =(A-B)(A+B)=7(A-B)=7
=> [4] 1=A-B
[3]+[4] => A=4 => 16=x+2 => x=14
[4]-[3] => B=3 => 9=x-5 => x=14
✓✓✓
It's a juicy problem....
🤗
this is a middle school level question in my country, I am now 20 and I learned this when I was 16, still remembering the best approach to solve this question just shows how goated my teacher is.... Love you Sir Navin Kumar
With the numbers offered this is a simple mental arithmetic problem. No pen or paper needed.
7 = (3 + 4) or (2 + 5) or (1 + 6)
I went for 3 + 4.
For the first square root to be 3 then X = 14 (14 - 5 = 9 and the square. root of 9 = 3)
For the second square root to be 4 then X = 14 (14 + 2 = 16 and the square root of 16 = 4)
Therefore X = 14
I did not bother checking the other two to see if they had a solution.
Same here. Ruled out complex numbers and went the way you did. No need to be inefficient with brain power.
@@afsoc4life when they set these types of questions you would hope they would make the problem meaningful in terms of the numbers they use that would justify the need for complexity.
I'm with you in this. X=14 by inspection. No pen or paper, 15 seconds but faster for people who think faster than me, which isn't hard. Same route, what adds to 7, can you make it happen, yes? Done. The only problem with 15 seconds is if you try 1+6 or 2+5 first, and then confirming they don't work and moving to 3+4 would easily push me past 15 seconds.
But yeah, let's make a 9 minute video about a 15 second problem 😂
correct, you used the 'look and see method'...
😇👍👍
🙏
Please do it in less steps. You dont have to write every obvious operation. This can be done in less than10 rows and will be still easy to follow.
I have 0 visualization skills. I need to see even the obvious steps written, or they slip away.
The steps are written to those who are not at a proficiency level, maybe a beginner or intermediate. Certainly, a proficient person like you, would think these steps are unnecessary.
You might understand it, but someone younger and earlier in math might not.
I think so too. When we solve such an equation, we use the concept called 自明解, which means evident solution. In this equation, we can clearly see one of the solutions is 14(this is 自明解), and then what we have to do is to proof that this is the only solution. This is easy because it is clear that when X increases, √X steadily increases. I don't know what this is called in English, but we call this 単調増加. 単調 means monotone and 増加 means increasing
I searched looked up these words in google. 自明解 is trivial solution and 単調増加 is monotonic increasing
Giải kiểu nầy sẽ bị trừ điểm vì thiếu điều kiện để phân thức trong căn tồn tại , nên ta phải đặt điều kiện ( Đk ) để phân thức trong căn tồn tại .
Đk : x-5>=0
X+2>=0
Vậy : x >= 5
...........
If it's a test, choices can help.
✌️
Graph on desmos. Answer in 30 sec x=14
It will be easy if you transfer one term from left to right. Then square both sides
👌
I just asked myself, "How do you get 7 from addition?" 7+0, 6+1, 5+2, and 4+3. And which are the closest together? 4+3. Then I squared both numbers to get 16 and 9. Then it was just simple math to get that 14-5=9 and 14+2=16, meaning x=14.
Sir, I dont get it
First you choose 4 and 3
In my head based on your choice
X - 5 = 16 , making x = 21
X + 2 = 9 , making x = 7
And is not the solution
Or everthing is based on luck ?
@erwinyonathan6348 Because you need to swap it. x-5=9, x+2=16. It only works if you subtract 5 to get the lower number and add 2 to get the higher one. You'll never get x if you subtract 5 to get the higher number and add 2 to get the lower one.
A=Sqrt（X+2）， B=Sqrt（X-5），A+B=7， A^2-B^2=（X+2）-（X-5）=7， A-B=1，A=4，B=3，X=B^2+5=3^2+5=14
Can you please explain more
How you get from A+B=7
To
A^2+B^2=(x+2)-(x-5)=7
I do not get it 😊
Your process of solving the equation matched with mine line by line. And i don't think this problem to be too complicated. It's too easy for a 6th grader.
@kamrunnahar3855 Yes, easy solution should be given to the students.
Isn't the left side a growing function? So the solution is just to find 1 answer. And 14 works. So why do we need 9 minutes to solve it?
At beginning, we must suppose that sqr (x - 5) >= 0 so x >= 5 in R.
You should considered that (a+b)²=a²+b²+2ab ins't always correct if you have a square root.
I have realized there are 3 main approaches
1. (a) Selecting the term / terms to be squared
(b) Making the adjustments for the term / terms to be squared
(c) Simplification
2. Use inspection to eliminate irrational solutions on the LHS : prior comment shows only 1 rational solution exists, the roots of the perfect squares ( 9, 16).
3. Reconstruct the method of problem construction : Abacus users have sight comparison tricks here.
(a) The LHS is the sum of 2 integers, and their sum is the RHS integer.
(b) The LHS is sum of the roots of 2 perfect squares.
(c) The difference between the 2 perfect squares is the difference of the 2 constants = 7.
(d) Investigate the perfect squares from 1 upwards
value square
1 1
2 4
3 9
4 16
5 25
The pair of perfect squares with a difference of 7 is (9, 16).
The sum of roots of the perfect squares is 3 + 4 = 7 (Value of RHS)
For the smaller term : root( x - 5), x - 5 = smaller perfect square (9). x = 14.
VERIFY
For the larger term : root( x + 2), root (14 + 2 = 16) = 4.
CONCLUSION
X = 14
NOTE
1. I don't doubt someone will say I have been very wordy, but my aim was clarity over brevity.
2. There was something about the problem which made me think an abacus user would have a huge advantage , even if they claimed they didn't.
3. The reconstructing the construction method should work for any integer power, but for odd integer powers, the negative values demand evaluation as well.
✌️
@@MasterTMathsClass
Thank you for the appreciation.
I realize now why I suspected an abacus was used in construction. The sum 7, the difference of the constants 7, sum to the answer 14. Very like the symmetry in a Japanese stone garden.
Chasing perfect squares in each radical, you quickly find x=14 as an obvious solution.
The function being increasing on its domain, the solution is unique.
You could also take the square on the left and right, move the non sqrt terms on one side, square again, develop and solve the quadratic.
Lots of work for nothin'
sqrt(x-5) = 7 - sqrt(x+2)
so x-5 = 49 - 14sqrt(x+2) + x + 2
so 56 = 14sqrt(x+2)
so 4 = sqrt(x+2)
so 16 = x + 2
so x = 14
Check
sqrt9 + sqrt16 = 3 + 4 = 7
✌️
How could I make 7 with easy root? 3 + 4? Let's try that. 9 and 16. 16 - 2 = 14 and 9 + 5 = 14. X =14.
But what if for some reason you choose
9-2=11 and 16+5=21
Than you will not find the value of x
So how do you sure 7 is 3+4 and not 4+3 ??
@erwinyonathan6348 because of thinking
I suspect the Japanese are playing a trick on everyone.
The result, RHS, is an integer 7.
The LHS is the sum of 2 roots. If the roots are irrational, the sum will be also irrational.
The possible integer sums adding up to 7 are
(a) 1,6
(b) 2,5
(c) 3, 4
(d) 4, 3
(e) 5, 2
(f) 6, 1
Considering only the 1st term of the LHS, the possible values of X are
(a) root(X - 5) = 1, squared = 1, giving X = 6.
(b) root(X - 5) = 2, squared = 4, giving X = 9.
(c) root(X - 5) = 3, squared = 9, giving X = 14.
(d) root(X - 5) =4, squared = 16, giving X = 21.
(e) root(X - 5) =5, squared =25, giving X = 30.
(f) root(X - 5) = 6, squared =36, giving X = 41.
Considering the 2nd term of the RHS, the valid value of X is
(a) root(6 + 2) which is root(8) = 2 root(2) which is irrational
(b) root(9 +2) which is root(11) which is irrational
(c) root(14+2) which is root(16) = 4, an integer value.
(d) root(21+2) which is root(23) which is irrational
(e) root(30+2) which is root(32) = 4 root(2) which is irrational
(f) root(41+2) which is root(43) which is irrational
Only X =14 has integer roots, 3+4 = 7, the solution is X = 14.
With the help of an abacus the possibilities can be explored in seconds.
I feel sure the solution I have outlined is a clumsy version of the numerical tricks an abacus user has available.
Note to self : Japanese Olympiad problem.
sqrt(x-5)+sqrt(x+2)=7
2x-3 +2sqrt(x-5)*sqrt(x+2)=49
sqrt(x-5)*sqrt(x+2)=26-x
{ (x-5)(x+2)= (26-x)^2 , 5
Square both sides:
x-5+x+2+2√(x²-3x-10)=49
2√(x²-3x-10)=-2x+52
√(x²-3x-10)=-x+26
x²-3x-10=x²-52x+676
49x=686
x=14
✌️
Assuming positive square root
by guesswork x=14 is a solution.
on the lhs we have a sum of two monotonically increasing functions.
lie the lhs is monotonically increasing function.
Thus, the is only one value of x equal to any given x
thus x=5 is the only real solution.
This is NOT Japanese. This is excruciating! I could have planned Operation Overlord twice on that much white board after solving this in x fewer steps
√x-5=7-√x+2
x-5=49+x+2-14√x+2
14√x+2=51+5
14√x+2=56
√x+2=4
x+2=16
x=14 ans 🇮🇳
✌️
x+2 and x-5 have 7 gaps.
Also, 16 and 9 have 7 gaps.
4 + 3 = 7
So x+2 = 16, x-5 = 9
x = 14
I solved in 10 seconds, X=14
There is a difference in getting the answer and solving it.
Avant tout debut vous avez oublié le domaine de definition
Thanks for thi video. From a rigorous point of view you may have precised the conditions of existence for x. Here x>=5, x
Nice... Thanks 😊
(x-5)^½+(x+2)^½=7
7-(x+2)^½=(x-5)^½
7-(x+2)^½=(x+2-7)^½
Let y=x+2
7-y^½=(y-7)^½
Let z=y^½
7-z=(z²-7)^½
z²-14z+49=z²-7
49-14z=-7
-14z=-56
z=4
y^½=4
y=16
x+2=16
x=14 ❤
Sum of two square roots is integer (7) so I guess two square roots are also integer and only 3+4=7.
✌️
Слева функция возрастающая, справа константа, значит, максимум один корень. Находим подбором.
Or you could just use the 'look and see method' and see x = 14 is a solution 😂😂
Please teach me this method 😂😂
Way do you solve this ? It's a simple question you can easily guess the answer 😂 if it was a test
Acá independiente de la manera que se resuelva, algunas mas cortas otras mas larga lo importante es usar la creatividad, yo lo hice de una manera mas corta usando variables auxiliares, de todas formas hay que agradecer lo que se expone en el vídeo, dejar los eggo a un lado, porque todo sirve incluyendo los comentarios, que también aportan al desarrollo
By inspection x = 14
✌️
Too lengthy.
You have solved in complicated way.
You made your solution too complicated and unnecessary. The problem is very simple. There are several such methods. One is simply bu transposing, you can write Sqrt(x-%) = 7- Sqrt(x+2). Now squaring both sides and simplyfying, you get x+2 = 16; which gives your answer: x = 14. Dr. Ajit Thakur (USA).
I subscribed to you, in case you ever want to make maths videos. Shorter methods help a lot for kids with ADHD.
Says the same guy that says the problem is simple, there is several such methods "/
Возводить в квадрат здесь не самый простой способ! Это у вас такой уровень образования в штатах, что доктор пишет полный бред?))
X-5=49-14{x+2}+x+2
14{x+2}=56
{x+2}=4
x+2=16
x=14
Your method not different with him.. Need square both side twice..
14 in a few seconds
✌️
X is 14
14,8
x = 14
let u=x-5 , --> x=u+5 , Vu+V(u+7)=7 , (V(u+7))^2=(7-Vu)^2 , u+7=49-14Vu+u , / -u , 14Vu=49-7 , 14Vu=42 , Vu=3 , u=9 ,
recall , x=u+5 , x=9+5 , x=14 , test , V(14-5)+V(14+2)=V9+V16 , --> 3+4=7 , OK ,
Nicely done
@@michaeledwards2251 Thanks!
14
Why did he make it so complicated
I don’t know, in Taiwan such a long writing means that you do math too little
Too much, not simple
Bull
14,8
14
✌️