Question

One positive integer is 1 less than twice another. The sum of their squares is 157. Find the integers.

Answer 1

• 11 and 6

Explanation:

Let the integers are x and y.

We have:

• x = 2y - 1
• x² + y² = 157

Substitute x into the second equation and solve for y:

• (2y - 1)² + y² = 157
• 4y² - 4y + 1 + y² = 157
• 5y² - 4y - 156 = 0
• D = (-4)² - 4*5*(-156) = 3136
• y = (4 ± √3136)/10 = (4 ± 56)/10
• y = 6 is the only integer solution

Find x:

• x = 2*6 - 1 = 11