Question 1

Find two consecutive positive even integers whose squares have the sum 340.

Solution:

Let x and (x + 2) are two positive even integers.

Sum of their squares = 340

x² + (x + 2)² = 340

x² + x² + 2 (x) (2) + 2² = 340

2 x² + 4 x + 4 - 340 = 0

2 x² + 4 x - 336 = 0

Now we are going to divide the whole equation by 2,so we get

2 x²/2 + 4 x/2 - 336/2 = 0/2

x² + 2 x - 168 = 0

x² - 12 x + 14 x - 168 = 0

x (x - 12) + 14 (x - 12) = 0

(x - 12) (x + 14) = 0

x - 12 = 0        x + 14 = 0

x = 12            x = -14

Since the required number is even positive integers we have to choose 12 only as answer.

Therefore two positive even inters are 12 and 14.

Verification:

Sum of their squares is 340

=  12² + 14²

=  144 + 196

=  340

Question 2

The sum of two squares of three consecutive natural numbers is 194. Determine the numbers.

Solution:

Let x , x + 1 and x + 2 are three consecutive natural numbers

Sum of their squares = 194

x² + (x + 1)² + (x + 2)² = 194

x² + x² + 1² + 2 (x) (1) + x² + 2² + 2(x) (2) = 194

x² + x² + 1 + 2 x + x² + 4 + 4 x = 194

x² + x² + x² + 2 x + 4 x + 1 + 4 = 194

3 x² + 6 x + 5 = 194

3 x² + 6 x + 5 - 194 = 0

3 x² + 6 x - 189  = 0

Now we are going to divide the whole equation by 3, So we will get

x² + 2 x - 63  = 0

x² + 9 x - 7 x - 63  = 0

x (x + 9) - 7 (x + 9)  = 0

(x + 9) (x - 7) = 0

x + 9 = 0          x - 7 = 0

x =  - 9            x = 7

Therefore the required numbers are 7,8,9

Verification:

The sum of two squares of three consecutive natural numbers is 194

=  7² + 8² + 9²

=  49 + 64 + 81

=  194