Question

Find two consecutive odd positive integers sum of whose squares is 970

Answer 1

Answer:

One number = 21

Another number = 23

Step-by-step explanation:

given that sum of square of two odd consecutive positive integers = 970

let one integer be x

so,

another odd integer = x + 2

given sum of square of the numbers

= 970

so,

According to the question,

x² + (x + 2)² = 970

x² + x² + 4x + 4 = 970

2x² + 4x + 4 - 970 = 0

2x² + 4x - 966 = 0

2x² + 46x - 42x - 966 = 0

2x(x + 23) -42(x + 23) = 0

(2x - 42)(x + 23) = 0

2x - 42 = 0

2x = 42

x = 42/2

x = 21

so,

one number will 21

another number = x + 2

= 21 + 2

= 23

so,

One number = 21

Another number = 23

Answer 2

Answer:

The required numbers = 21 and 23

Step-by-step explanation:

let the integer be x and x + 2

given sum of squares = 970

so,

ATQ,

x² + (x + 2)² = 970

x² + x² + 4x + 4 = 970

2x² + 4x + 4 - 970

2x² + 4x - 966 = 0

2x² + 46x - 42x - 966 = 0

2x(x + 23) -42(x + 23) = 0

(2x - 42)(x + 23) = 0

x + 23 = 0

x = -23

2x - 42 = 0

2x = 42

x = 42/2

x = 21

so,

one number will 21

because the number must be positive

and

another number = 21 + 2

= 23

so,

The required numbers are

21 and 23