Question

sum of two numbers is 36 and there product is 128

Answer 1

Hello!

let the first number be x
therefore the other number =36-x

by the problem,
x (36-x)=128
=>x^2-36x+128=0
=>x^2-32x-4x+128=0
=>x (x-32)-4 (x-32)=0
=>(x-4)(x-32)=0
therefore,
x-4=0 or, x-32=0
=>x=4 =>x=32

the two numbers are 4 and 32.

hope it helps... :)

Answer 2

hi Rajat

let the two numbers br x and y respectively.

And,
According to question

x + y = 36 ......... (1)

and
xy = 128

x = 128/ y
putting X = 128/y in equation (1), we get

128/y + y = 36
$128 + {y}^{2} = 36y$
${y}^{2} - 36y + 128 = 0$
now we need to factorise the obtained equation.

${y}^{2} - 32y - 4y + 128 = 0 \\ y(y - 32) - 4(y - 32) = 0 \\ (y - 4)(y - 32) = 0 \\ y = 4 \: or \: y = 32$
When y = 4
x = 32

and when y = 32
x = 4



hence the required number could be either (32,4) or (4,32)


hope it helps
#jerri