The sum of two numbers is 48 and their product is 432. Find the numbers.

Solution:

Let first numbers be $x$ and other $(48-x)$

Then according to question

$x(48-x)=432$

$48 x-x^{2}=432$

$x^{2}-48 x+432=0$

$x^{2}-36 x-12 x+432=0$

$x(x-36)-12(x-36)=0$

 

$(x-36)(x-12)=0$

$(x-36)=0$

$x=36$

Or

$(x-12)=0$

$x=12$

Thus, two number be 36,12