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Question
Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.
Solution
Let the larger and smaller parts be x and y, respectively.
According to the question :
`x+y=16` ...........(1)
`2x^2=y^2+164` ..............(2)
From (i), we get:
`x=16-y` ............(3)
From (ii) and (iii), we get:
`2(16-y)^2=y^2+164`
⇒`2(256-32y+y^2)=y^2+164`
⇒`512-64y+2y^2=y^2+164`
⇒`y^2-64y+348=0`
⇒`y^2-(58+6)y+348=0`
⇒`y^2-58y-6y+348=0`
⇒`y(y-58)(y-6)=0`
⇒`y-58=0 or y-6=0`
⇒`y=6 (∵y<16)`
Putting the value of y in equation (3), we get
`x=16-6=10`
Hence, the two natural numbers are 6 and 10.
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