Question

two positive numbers differ by 5 and sum of their square is 433 find the number​

Answer 1

I am not sure about this answer.

Hope it will help you.

Answer 2

Answer:

The required numbers are 12 & 17.

Step-by-step-explanation:

Let the greater number be x.

And the smaller number be y.

From the first condition,

x - y = 5

⇒ x = 5 + y

⇒ x = y + 5 - - - ( 1 )

From the second condition,

x² + y² = 433

⇒ ( y + 5 )² + y² = 433 - - - [ From ( 1 ) ]

⇒ y² + 2 * y * 5 + 5² + y² = 433

⇒ y² + y² + 10y + 25 = 433

⇒ 2y² + 10y + 25 = 433

⇒ 2y² + 10y + 25 - 433 = 0

⇒ 2y² + 10y - 408 = 0

⇒ 2y² + 34y - 24y - 408 = 0

⇒ 2y ( y + 17 ) - 24 ( y + 17 ) = 0

⇒ ( y + 17 ) ( 2y - 24 ) = 0

⇒ ( y + 17 ) = 0 OR ( 2y - 24 ) = 0

⇒ y + 17 = 0 OR 2y - 24 = 0

⇒ y = - 17 OR 2y = 24

⇒ y = - 17 OR y = 24 ÷ 2

⇒ y = - 17 OR y = 12

But the given numbers are positive.

∴ y = - 17 is unacceptable.

∴ y = 12

∴ Smaller number = 12

By substituting y = 12 in equation ( 1 ), we get,

x = y + 5 - - - ( 1 )

⇒ x = 12 + 5

⇒ x = 17

∴ Greater number = 17

∴ The required numbers are 12 & 17.