Question

if x=7-2√10/3 and y=7+2√10/3 find the value of x^2+xy+y^2​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

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Step-by-step explanation:

$so \: here \\ x = \frac{7 - 2 \sqrt{10} }{3} \\ y = \frac{7 + 2 \sqrt{10} }{3} \\ \\ so \: thus \: then \\ xy = \frac{7 - 2 \sqrt{10} }{3} \times \frac{7 + 2 \sqrt{10} }{3} \\ = \frac{(7 - 2 \sqrt{10})(7 + 2 \sqrt{10} }{9} \\ = \frac{49 - (4 \times 10)}{9} \\ = \frac{49 - 40 }{9} \\ \\ = \frac{9}{9} \\ ie \: \: xy = 1$

$now \: using \\ x {}^{2} + y {}^{2} + xy \\ = ( \frac{7 - 2 \sqrt{10} }{3} ) {}^{2} + 1 + ( \frac{7 + 2 \sqrt{10} }{3} ) {}^{2} \\ \\ = \frac{49 + 40 - 28 \sqrt{10} }{9} + \frac{49 + 40 + 28 \sqrt{10} }{9} + 1 \\ \\ = \frac{49 + 40 - 28 \sqrt{10 } + 49 + 40 + 28 \sqrt{10} }{9} + 1 \\ \\ = \frac{89 + 89}{9} + 1 \\ \\ = \frac{178 + 9}{9} \\ \\ = \frac{187}{9}$