Question

√(x² - x - 2) = 2x-4

Answer 1

Ответ:

$2;\ 3$

Объяснение:

ОДЗ:

1)

$x^2-x-2\ge 0$

Нули:

$x^{2} - x - 2 =0\\\\ a=1 ,\ \ b=-1 ,\ \ c=-2\\\\ D = b^2 - 4ac = ( - 1)^2 - 4\cdot1\cdot( - 2) = 1 + 8 = 9\\\\\sqrt{D} =\sqrt{9} = 3\\\\ x_1=\frac{-b-\sqrt{D}}{2a}=\frac{1-3}{2\cdot1}=\frac{-2 }{2 }=-1\\\\ x_2=\frac{-b+\sqrt{D}}{2a}=\frac{1+3}{2\cdot1}=\frac{4}{2}=2 \\\\x\in \left( -\infty;-1 \right]\cup\left[ 2 ;+\infty\right)$

2)

$2x-4\ge 0\\\\2x\ge 4\ \ \ \ |:2\\\\x\ge2\\\\x\in[2;+\infty)$

С 1) и 2)

$\underline{x\in[2;+\infty)}$

$\sqrt{x^2-x-2} = 2x - 4\ \ \ |()^2\\\\x^2-x-2=4x^2-16x+16\\\\x^2-x-2-4x^2+16x-16=0\\\\-3x^2+15x-18=0\ \ \ |:(-3)\\\\x^2-5x+6=0a=1 ,\ \ b=-5 ,\ \ c=6\\\\ D = b^2 - 4ac = ( - 5)^2 - 4\cdot1\cdot6 = 25 - 24 = 1\\\\\sqrt{D} =\sqrt{1} = 1\\\\ x_1=\frac{-b-\sqrt{D}}{2a}=\frac{5-1}{2\cdot1}=\frac{4 }{2 }=2\\\\ x_2=\frac{-b+\sqrt{D}}{2a}=\frac{5+1}{2\cdot1}=\frac{6}{2}=3$