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.
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
(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 ❤
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
✌️
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.
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 😊
Why did he make it so complicated
I don’t know, in Taiwan such a long writing means that you do math too little
I solved in 10 seconds, X=14
14,8
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 ,
X= 14. No necesito tu explicación, ya lo aprendí en la escuela pública.
14,8