Question

the product of two numbers if 45 and the sum if their squares is 106. find the numbers.

Answer 1

Hi,

Let m , n are two numbers ,

according to the problem given ,

mn = 45 --( 1 )

m² + n² = 106 ---( 2 )

( m + n )² = m² + n² + 2mn

= 106 + 2 × 45

= 106 + 90

= 196

m + n = √196

m + n = 14 ---( 3 )

( m - n )² = m² + n² - 2mn

= 106 - 2 × 45

= 106 - 90

= 16

m - n = √ 16

m - n = 4 ---( 4 )

add equation ( 3 ) and ( 4 ) , we get

2m = 18

m = 18/2

m = 9

put m = 9 in equation ( 3 ) , we get

9 + n = 14

n = 14 - 9

n = 5

Therefore ,

Required two numbers are ,

m = 9 ,

n = 5

I hope this helps you.

: )

n =

Answer 2

HELLO........FRIEND!!

THE ANSWER IS HERE,

let the two numbers be x and y.

From the question,

=> xy =45........(.1)

$= > \: {x}^{2} + {y}^{2} = 106............(2)$

Multiply eq (1) with 2.

=> 2xy = 90......... (3)

Adding eq (2) & (3).


$= > \: {x}^{2} + {y}^{2} + 2xy = 106 + 90$


$= > \: {(x + y)}^{2} = 196$

$= > \: x + y = 14.........(4)$

Now subtracting the eq(3) from (2).

$= > \: {x}^{2} + {y}^{2} - 2xy = 106 - 90$

$= > {(x - y)}^{2} = 16$

$= > \: x - y = 4............(5)$


Adding eq (4) & (5)

=> x+y+x-y = 14+4

=> 2x= 18.

=> x =9.

Substitute the value of x in eq(4).

=> 9 +y = 14.

=> y = 14-9.

=> y = 5.


Hence, The two numbers are 9 and 5.

:-)Hope it helps u.