Question

the difference between the squares of two numbers is 3. Twice the square of the first number increased by the square of the second number is 9. find the numbers.

Answer 1

Answer:

The number are $\sqrt{6}$ and $\sqrt{6}$ .

Step-by-step explanation:

Given as :

The difference between the squares of two number = 3

And , Twice the square of the first number increased by the square of the second number is 9

Now, Let The first number = F

And The second number = S

So , According to question

F² - S² = 3                  .......A

2 F² - S² = 9               ........B

Now, solving the equations

( 2 F² - S² ) - ( F² - S² ) = 9 - 3

Or, ( 2 F² - F² ) + ( S² - S² ) = 6

Or, F² + 0 = 6

∴   F = $\sqrt{6}$

Putting the value of F in eq A

So, (  $\sqrt{6}$ )² - S² = 3  

Or, 6 - 3 = S²

Or, S² = 3

∴   S = $\sqrt{3}$

So, The number are $\sqrt{6}$ , $\sqrt{6}$

Hence The number are $\sqrt{6}$ and $\sqrt{6}$ . Answer