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Question
Divide 57 into two parts whose product is 680.
Solution
Let the two parts be x and (57-x)
According to the given condition,
`x(57-x)=680`
⇒`57x-x^2=680`
⇒`x^2-57x+680=0`
⇒`x^2-40x-17x+680=0`
⇒`x(x-40)-17(x-40)=0`
⇒`(x-40) (x-17)=0`
⇒`x-40=0 or x-17=0`
⇒`x=40 or x=17`
When `x=40`
`57-x=57-40=17`
When` x=17`
`57-x=57-17=40`
Hence, the required parts are 17 and 40.
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