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Same idea in slightly different notations: Let S=sum of a, b, c & d and let Q=sum of their squares. (Note that Q=10x - 16 as in the video). Then 2+Q/2 = S ≤ 4√(Q/4) where the "=" sign is shown in the vidoe but the "≤" sign comes from the QM-AM inequality. This solves to (√Q-2)² ≤ 0 , i.e., Q=4. In terms of x, that is Q = 10x -16 =4 and thus x = 2. Essentially, this approach completes one square in terms of Q, compared with completing 4 squares in terms of a, b, c, and d.
a =√(x - 1), b =√(2 x -3), c = √(3 x -5) d = √ (4 x - 7) Hereby a^2 + b^2 + c^2 + d^2 = 10 x - 16 2( a + b + c + d) = a^2 + b^2 + c^2 + d^2 + 4 (a - 1)^2 + (b - 1)^2 + (c - 1)^2 + (d - 1)^2 = 0 Hereby x - 1 = 2 x - 3 = 3 x - 5 = 4 x - 7 = 1 all 4 equations give same solution x = 2
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Same idea in slightly different notations: Let S=sum of a, b, c & d and let Q=sum of their squares. (Note that Q=10x - 16 as in the video). Then
2+Q/2 = S ≤ 4√(Q/4)
where the "=" sign is shown in the vidoe but the "≤" sign comes from the QM-AM inequality.
This solves to (√Q-2)² ≤ 0 , i.e., Q=4.
In terms of x, that is Q = 10x -16 =4 and thus x = 2.
Essentially, this approach completes one square in terms of Q, compared with completing 4 squares in terms of a, b, c, and d.
Nice method and thanks for sharing!
x-1=(x-2)+1,2x-3=2(x-2)+1,3x-5=3(x-2)+1,4x-7=4(x-2)+1
5x-6=5(x-2)+4
Here when x=2,lhs=√1+√1+√1/√1=4=rhs
So x=2
Thanks for sharing
Solution by insight
1+1+1+1=5×2-6
x=2
Nice try!
a =√(x - 1), b =√(2 x -3), c = √(3 x -5)
d = √ (4 x - 7)
Hereby
a^2 + b^2 + c^2 + d^2 = 10 x - 16
2( a + b + c + d)
= a^2 + b^2 + c^2 + d^2 + 4
(a - 1)^2 + (b - 1)^2
+ (c - 1)^2 + (d - 1)^2 = 0
Hereby
x - 1 = 2 x - 3 = 3 x - 5 = 4 x - 7 = 1
all 4 equations give same solution
x = 2
Thanks for sharing
😂🎉
Thanks