Question

The sum of natural number and its positive square root is 132. Find the number. 

 

Answer 1

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.