Question

Given x² = 17x +y and y²= x+17y . Find :
$\sqrt{ {x}^{2} + {y}^{2} + 1 }$
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Answer 1

Answer:

Step-by-step 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

Answer:

${x}^{2} = 17x + y....(1) \\ {y}^{2} = x + 17y....(2) \\ (1) - (2) \\ {x}^{2} - {y}^{2} = 16x - 16y \\ {x}^{2} - {y}^{2} = 16(x - y) \\ (x - y)(x + y) = 16(x - y) \\ (x - y) \neq0 \\ x + y = 16 ....(3)\: since \: x \neq \: y \\ (1) + (2) \\ {x}^{2} + {y}^{2} = 18x + 18y \\ {x}^{2} + {y}^{2} = 18(x + y) \\ subtitute \: (3) \\ {x}^{2} + {y}^{2} = 18 \times 16 \\ {x}^{2} + {y}^{2} = (17 + 1)(17 - 1) = 289 - 1 \\ {x}^{2} + {y}^{2} + 1 = 289) \: take \: sqr \\ \sqrt{ {x}^{2} + {y}^{2} + 1} = \sqrt{289} \\ \therefore \sqrt{ {x}^{2} + {y}^{2} + 1} = 17$