Question

решите пожалуйста
$2 \sqrt{5 - {x}^{2} } = x - 1$

Answer 1

$2 \sqrt{5-x^2} =x-1 \\ \sqrt{4(5-x^2)} =x-1 \\ \sqrt{(20-4x^2)} = x-1 \\ 20-4x^2=(x-1)^2 \\ -4x^2+20=x^2-2x+1 \\ -5x^2+2x+19=0 \\ 5x^2-2x-19=0 \\ D=b^2-4ac = (-2)^2-4*5*(-19)= \sqrt{384} = 8 \sqrt{6} \\ x1= \frac{-b+ \sqrt{D} }{2a} = \frac{2+ 8 \sqrt{6} }{10} = \frac{1}{5} + \frac{4 \sqrt{6} }{5} \\ x2= \frac{-b- \sqrt{D} }{2a} = \frac{2- 8 \sqrt{6} }{10} = \frac{1}{5} - \frac{4 \sqrt{6} }{5}$