Question

if sum and product of two numbers are 24 and 128 respectively find the numbers

Answer 1

x+y=24            -(1)

xy=128             -(2)

From(1),

x=24-y

Put value of x in (2)

Therefore,

y(24-y)=128

24y-y^{2}=128

y^{2}-24y+128=0

y^{2}-16y-8y+128=0

(y-16)(y-8)=0

Therefore,

y=16 or y=8

If y=8,

then from (1),x=24-8=16

Ify=16

then from(1),x=24-16=8

Answer 2

Given: Sum of two numbers = 24

           Product of those numbers = 128

To find: The numbers

Let: The numbers be $x$ and $y$

Solution: According to the given question and assumption made:

$x + y = 24$     ...(1)     or     $x = 24 -y$

$xy = 128$        ...(2)

Putting value of  $x$ from equation (1) in equation (2)

⇒ $(24 - y )y = 128$

⇒ $24y - y^{2} = 128$

⇒ $y^{2} - 24y + 128 = 0$

Expanding the equation using middle term splitting method:

⇒ $y^{2} - 16y - 8y + 128 = 0$

⇒ $y(y - 16) - 8(y - 16) = 0$

⇒ $(y - 8)(y - 16) = 0$

When $(y-8) = 0, y = 8$

Putting $y= 8$  in equation (1)

⇒ $x + 8 = 24$

⇒ $x = 24 - 8 = 16$

When $(y-16)=0, y= 16$

Putting $y = 16$ in equation (1)

⇒ $x + 16 = 24$

⇒ $x = 24 - 16 = 8$

Hence, the numbers are (16, 8) or (8, 16).