One positive integer is 6 less than twice another. The sum of their squares is 680 . Find the integers.

Answer 1

a² + (2a - 6)² = 680

a² + 4a² -- 24a + 36 = 680

5a² - 24a - 644 = 0

5a² - 70a + 46a - 644 = 0

5a(a - 14) + 46(a - 14) = 0

(5a + 46)(a - 14) = 0

5a + 46 = 0

5a = -46

a = -46/5 ------> not positive or an integer

a - 14 = 0

a = 14

2(14) - 6 = 22

14 and 22

Answer 2

x = 2y-6

x^2 + y^2 = 680

(2y-6)^2 +y^2 = 680

4y^2 -24y +36 +y^2 = 680

5y^2 -24y - 644 = 0

y = 24/10 +/-.1sqr(24^2 +20(644))

y = 2.4 +/-.1sqr(576 +12880)

y = 2.4 +/-.1sqr13456

y= 2.4 +/- .11.6

y = 14 (ignore the negative square root)

x= 2y-6 = 28-6= 22

check the solution

22^2 + 14^2

484 + 196

= 680

Answer 3

x² + (2x - 6)² = 680

x² + 4x² - 24x + 36 = 680

5x² - 24x - 644 = 0

Using the quadratic formula, x = 14. So, 2x - 6 = 22.

The integers are 14 and 22.