Question

2x=√4-x+x² решите уравнение

Answer 1

Ответ:

1; -1 1/3

Объяснение:

$\sqrt{4-x+x^{2} } =2x\\ (\sqrt{4-x+x^{2} })^{2}=(2x)^{2}\\ |\sqrt{4-x+x^{2} }|  =(2x)^{2}\\4-x+x^{2} =4x^{2} \\ x^{2} -4x^{2} -x+4=0\\-3x^{2} -x+4=0/*(-1)\\3x^{2} +x-4=0\\D=b^{2} -4ac=1+48=49\\ \sqrt{D} =7\\ x_{1} =\frac{-b-\sqrt{D} }{2a} =\frac{-1-7}{2*3}=\frac{-8}{6}  =-\frac{4}{3} =-1\frac{1}{3} \\ x_{2} =\frac{-b+\sqrt{D} }{2a} =\frac{-1+7}{2*3}=\frac{6}{6} =1$

проверка:

при х= -1 1/3

$-1\frac{1}{3} =-\frac{4}{3}\\\\ 4-x+x^{2} =4x^{2}\\ 4-(-\frac{4}{3} )+(\frac{4}{3} )^{2} =4+\frac{4}{3} +\frac{16}{9} =4+\frac{4*3}{3*3} +\frac{16}{9} =4+\frac{12}{9}+\frac{16}{9}  =\\ 4+\frac{28}{9} =4+3\frac{1}{9}=7\frac{1}{9} \\\\ 4*(\frac{4}{3}) ^{2} =4*\frac{16}{9} =\frac{64}{9}=7\frac{1}{9}\\ \\ 7\frac{1}{9}=7\frac{1}{9}$

при х=1

$4-x+x^{2} =4x^{2} \\ 4-1+1^{2}=4-1+1=4+0=4\\4*1^{2}=4*1=4\\4=4$