Question

if a positive number exceeds its positive square roots by 12,then find the number.

Answer 1

let the positive number be x 
according to question 
√x + 12 = x 
⇒√x= x - 12
squaring both sides
⇒x= x² - 24x +144
⇒x² - 25x + 144 = 0 
⇒x² - 16x - 9x +144=0
⇒x(x - 16) - 9(x - 16)=0
⇒(x - 9)(x - 16)=0
so the number is 9,16

Answer 2

consider the number is x
a/q,
          x = √x+12
⇒       12-x = -√x
on squaring both side of the equation ,    
$12^{2} -2.12.x+ x^{2} =(- \sqrt{x}) ^{2} \\ x^{2} -24x+144=x \\ x^{2} -24x-x+144=0 \\ x^{2} -25x+144=0 \\ x^{2} -16x-9x+144=0 \\ x(x-16)-9(x-16)=0 \\ (x-16)(x-9)=0$
so,
                x-16 = 0 ⇒ x = 16
     and     x-9 = 0 ⇒ x= 9 
therefore numbers will be 9,16