Question

2 numbers whose product is 132 and diffrrence is 1

Answer 1

Let X and Y be the Numbers(Where X>Y)

So According to the Question..

XY=132→(1)
And
X-Y=1
→X=1+Y→(2)

Substituting (2) in (1)
→(1+Y)Y=132
→Y+Y²=132
→Y²+Y-132=0

Here,
D=b²-4ac=1-(-132×4)
=1+528
=529.

now
$y = \frac{ - b( + -) ( \sqrt{d} }{2a} \\ = \frac{- 1( + - )( \sqrt{529} }{2 \times 1} \\ = \frac{ - 1( + - )(23)}{2}$
So Y=(-1+29)/2=28/2=14

or Y=(-1-29)/2=(-30)/2=(-15)

When Y=14, X=15 { 1+Y [eq(2)] }
When Y=-15, X=-14 { 1+Y [eq(2)] }

So The Required No: are ±(15and14)

Hope it Helps...
Regards,
Leukonov.