Two positive numbers differ by 5 and the sum of their squares is 193 . find the two numbers Question Two positive numbers differ by 5 and the sum of their squares is 193 . find the two numbers in progress 0 Math Maya 3 weeks 2021-10-04T15:03:13+00:00 2021-10-04T15:03:13+00:00 2 Answers 0 views 0

## Answers ( )

## The two numbers are 7 and 12.

Step-by-step explanation:Let x and y be the two numbers.

Then it is given that x = y + 5 ————(1)

x² + y² = 193 ————-(2)

Substituting (1) in (2), we get:

(y + 5)² + y² = 193

25 + y² + 10y + y ² = 193

2y² + 10y – 168 = 0

Let’s solve the quadratic equation.

y² + 5y – 84 = 0

y² + 12y – 7y – 84 = 0

y (y + 12) – 7(y + 12)

(y + 12) (y – 7) = 0

Therefore

y = -12 or +7It is mentioned that the numbers are positive whole numbers. So y can only be +7.

Substituting

y = 7in equation 1, we get:x = 12So

the two numbers are 7 and 12.Answer:## Let the two positive whole numbers be x and y.

## Given that these numbers differ by = 5

## Therefore x – y = 5

## x = 5 + y

## Also given that the sum of their squares = 193

## Therefore

## x² + y² = 193

## Substituting the value of x

## (5 + y)² + y² = 193

## 25 + y² + 10y + y ² = 193

## 2y² + 10y – 168 = 0

## y² + 5y – 84 = 0

## y² + 12y – 7y – 84 = 0

## y(y + 12) – 7(y + 12)

## (y + 12) (y – 7) = 0

## y = -12 or y = 7

## Rejecting y = -12 as given they are positive whole numbers.

## Therefore y = 7

## x – 7 = 5

## x = 12

## The two numbers are 12 and 7.

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