Question

Sum and product of two numbers are 24 and 128 respectively

Answer 1

let x and y be the two numbers.

so,
x + y = 24
x = 24 - y → (1)

and,
xy = 128
(24 - y) y = 128
24y - y² = 128
y² - 24y + 128 = 0
y² - 16y - 8y + 128 = 0
y (y - 16) - 8 (y - 16) =0
(y - 8)(y - 16) = 0
so,
y = 8 or y = 16

put y in (1)
so,
x = 8 or x = 16

so,
the numbers are 8 and 16

Answer 2

Answer:

Two numbers are (16,8 )or (8,16)

Step-by-step explanation:

Given;

Sum of two no's = 24

Let one no = x

∴Other no. = 24-x

Product of two no's = 128

⇒x(24 - x)= 128

⇒24x-$x^{2}$-128 =0

$x^{2}$- 24x+128 =0 (rearrange the terms and multiply both side by negative sign)

$x^{2} -16x-8x+128=0$  (by factorising method)

x(x- 16) - 8(x -16)=0

 (x-16 ) (x-8)=0

 Either x-16=0 or (x-8)=0

       x= 16 or 8

When x= 16 ,other no.=24-16=8

When x= 8, other no. = 24- 8=16