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Question
The sum of natural number and its positive square root is 132. Find the number.
Solution
Let the required natural number be x.
According to the given condition,
`x+sqrtx=132`
Putting sqrtx=y or x=y^2, we get
`y^2+y=132`
⇒`y^2+y-132=0`
⇒`y^2+12y-11y-132=0`
⇒`y(y+12)-11(y+12)=0`
⇒`(y+12)(y-11)=0`
⇒`y+12=0 or y-11=0`
⇒`y=-12 or y=11`
∴ `y=11` ( y cannot be negative)
Now,
`sqrtx=11`
⇒` x=(11)^2=121`
Hence, the required natural number is 121.
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