Question

teorema de pitagoras con expresion algebraica
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Answer 1

Explicación paso a paso:

$c {}^{2} = a {}^{2} + b {}^{2} \\ c = 9 \\ a = x + 2 \\ b = 2x - 1 \\ \\ 9 {}^{2} = (x + 2) {}^{2} + (2x - 1) {}^{2} \\ 81 = (x {}^{2} + 4x + 4) + (4x {}^{2} - 4x + 1) \\ 81 = x {}^{2} + 4x + 4 + 4x {}^{2} - 4x + 1 \\ 81 = 5x {}^{2} + 5 \\ 5x {}^{2} = 81 - 5 \\ 5x {}^{2} = 76 \\ x {}^{2} = \frac{76}{5} \\ x {}^{2} = 15.2 \\ x = \sqrt{15.2 } \\ x = 3.9$

$a = x +2 \: \: \: \: \: \: b = 2x - 1\\ a = 3.9 + 2 \: \: \: \: \: \: b = 2(3.9) - 1 \\ a = 5.9 \: \: \: \: \: \: b = 7.8 - 1 \\ \: \: \: \: \: \: \: \: \: \: \: b = 6.8$