Question

Given x² = 17x +y and y²= x+17y . Find :

`\sqrt{ {x}^(2) + {y}^(2) + 1 }`
​

Answer 1

Explanation:

x² = 17x + y

y² = x + 17y

x² + y² = 17x + y + x + 17y

x² + y² = 18x + 18y

x² + y² + 1 = 9* 2(x + y) + 1

√(x² + y² + 1) = 3√(2x + 2y + 1)

Answer 2

`{x}^(2) = 17x + y....(1) \\ {y}^(2) = x + 17y....(2) \\ (1) - (2) \\ {x}^(2) - {y}^(2) = 16x - 16y \\ {x}^(2) - {y}^(2) = 16(x - y) \\ (x - y)(x + y) = 16(x - y) \\ (x - y) \\eq0 \\ x + y = 16 ....(3)\: since \: x \\eq \: y \\ (1) + (2) \\ {x}^(2) + {y}^(2) = 18x + 18y \\ {x}^(2) + {y}^(2) = 18(x + y) \\ subtitute \: (3) \\ {x}^(2) + {y}^(2) = 18 * 16 \\ {x}^(2) + {y}^(2) = (17 + 1)(17 - 1) = 289 - 1 \\ {x}^(2) + {y}^(2) + 1 = 289) \: take \: sqr \\ \sqrt{ {x}^(2) + {y}^(2) + 1} = √(289) \\ \therefore \sqrt{ {x}^(2) + {y}^(2) + 1} = 17`