Question

the sum of the squares of two consecutive even numbers is 6500.which is the smaller number? ​

Answer 1

56 is the answer.

let the two numbers be x and x+2

squaring them,

x^2± (x+2)^2=6500

x^2+x^2+4x+4=6500

2x^2+4x+4=6500

2x^2+4x-6496=0

x^2+2x-3248=0

x^2+58x-56x-3248=0

(x+58)(x-56)=0

x=56.

x+2=56+2=58

therefore, the smallest number is 56.

Answer 2

Answer:

The smaller no is 56.

Step-by-step explanation:

Let the two consecutive even numbers be x and (x+2).

According to the question,

x² + (x+2)² = 6500

or, x² + (x² + 2 × x × 2 + 2²) = 6500 (using formula)

or, x² + x² + 4x + 4 = 6500

or, 2x² + 4x = 6496 (calculation)

or, 2(x² + 2x) = 6496

or, x² + 2x = 3248

or, x² + 2x - 3248 = 0

or, x² + (58 - 56)x - 3248 = 0 (Mid term break)

or, x² + 58x - 56x - 3248 = 0

or, x(x + 58) - 56(x + 58) = 0

or, (x + 58) (x - 56) = 0

Either,

x + 58 = 0

Therefore, x = -58 (rejected)

Or,

x - 56 = 0

Therefore, x = 56

Again,

x + 2 = 56 + 2

        = 58

Therefore, The smaller number is 56.