Question

find two positive number whose squares have the difference 48 and the sum of number is 12​

Answer 1

Let one of the number be x

Since,

Their sum is 12,the other number would be 12-x

ATQ,

(12-x)² - x²=48

→144-2x+x²-x²=48

→144-2x=48

→2x=96

→x=48

The other number would be:

12-x

=12-48

= -36

Hence,the numbers are 48 and -36

Answer 2

Step-by-step explanation:

Hi,

Let Two positive number be x and y .

According to the question,

( x)² - (y)² = 48----------(1)

And,

x + y = 12 -----------(2)

From equation (1) , we get

x + y = 12

x = ( 12 - y ) ---------(3)

Putting the value of x in equation (1), we get

(x)² - (y)² = 48

( 12 - y )² - (y)²= 48

(12)² + (y)² - 2 * 12 * y - y² = 48

144 + y² - 24y - y² = 48

-24y + 144 = 48

-24y = 48 - 144

-24y = -96

Y = 96/24

Y = 4

Putting the value of y in equation (3) , we get

x = 12 - y = 12 - 4 = 8

Hence,

Two positive numbers are 4 and 8

Hope it will help you :)