Question

the two natural number which differ by 3 and the sum of whose squares is 117 is

Answer 1

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Answer 2

Hey


Let the two numbers be x and y respectively .

They differ from 3 .

So ,

x - y = 3

=> x = y + 3 –––( i )


And sum of their squares is 117 .

So ,

x² + y² = 117

Now , putting the value of x from eq ( i )

So ,


( y + 3 ) ² + y² = 117

=> y² + 9 + 6y + y² = 117

=> 2y² + 6y + 9 - 117 = 0

=> 2y² + 6y - 108 = 0

=> 2 ( y² + 3y - 54 ) = 0

=> y² + 3y - 54 =0

=> y² + 9y - 6y - 54 = 0

=> y ( y + 9 ) - 6 ( y + 9 ) = 0

=> ( y - 6 ) ( y + 9 ) = 0

So , when

( y - 6 ) =0

=> y = 6 .

So , x = y + 3

=> x = 6 + 3

=> x = 9

And when ,

( y + 9 ) = 0

=> y = -9

So

x = y + 3

=> x = (-9) + 3

=> x = -6 .

Both you can take ( positive value and negative value ) .


thanks :)