SOLUTION: I don't understand this problem. The square of a number exceeds that number by twelve.Find the number(two answers).

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Question 415129: I don't understand this problem.
The square of a number exceeds that number by twelve.Find the number(two answers).
Found 2 solutions by mananth, sudhanshu_kmr:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
let number be x
x^2-x=12
x^2-x-12=0
x^2-4x+3x-12=0
x(x-4)+3(x-4)=0
(x+3)(x-4)=0
x=-3 OR x=4

Answer by sudhanshu_kmr(1152) About Me  (Show Source):
You can put this solution on YOUR website!

let number is x, then square of it = x^2

x^2 = x+12
=> x^2 -x -12 =0
=> x^2 -4x+ 3x -12 =0
=> x(x-4) +3(X -4) = 0
=>(X-4)(x+3 ) = 0

so, x = 4 , x= -3
if number is positive then only x=4,
otherwise x= -3 and 4 also because -3 is also a number and difference of -3 and
its square i.e 9 is 12.