Question

Divide $12$ into two parts such that their product is $32$.

Answer 1

We have,

Let the two number is x and y.

So,

$x+y=12......\left(1\right)$

Now, According to given question,

$xy=32$

$y=\frac{32}{x}......\left(2\right)$

From equation ( 1) and ( 2) to, and we get,

$x+\frac{32}{x}=12$

$\Rightarrow x^2+32=12x$

$\Rightarrow x^2-12x+32=0$

$\Rightarrow x^2-(8+4)x+32=0$

$\Rightarrow x^2-8x-4x+32=0$

$\Rightarrow x(x-8)-4(x-8)=0$

$\Rightarrow (x-8)(x-4)=0$

If, $x-8=0,x=8$

If,$x-4=0,x=4$

So,

From ( 1) to,

$x+y=12$

$\Rightarrow 8+y=12$

$\Rightarrow y=4$

Hence, the number is $8and4$.