Question

Иррациональное уравнение $3\sqrt{3x^2+2x-4}-2x=3x^2-2$

Answer 1

$3\sqrt{3x^{2}+2x-4 }-2x=3x^{2} -2\\\\3\sqrt{3x^{2}+2x-4 }=(3x^{2} +2x-4)+2$

Сделаем замену :

$\sqrt{3x^{2}+2x-4 } =m,m\geq 0 \ \Rightarrow \ 3x^{2}+2x-4=m^{2}\\\\3m=m^{2}+2\\\\m^{2}-3m+2=0\\\\m_{1} =1\\\\m_{2}=2- \ teorema \ Vieta\\\\1)m=1\\\\\sqrt{3x^{2}+2x-4 }=1\\\\(\sqrt{3x^{2}+2x-4 })^{2} =1^{2}\\\\3x^{2}+2x-4=1\\\\3x^{2}+2x-5=0\\\\D=2^{2}-4*3*(-5)=4+60=64=8^{2}\\\\x_{1}=\frac{-2-8}{6}=-\frac{10}{6}=-1\frac{2}{3} \\\\x_{2}=\frac{-2+8}{6} =1\\\\2)m=2\\\\\sqrt{3x^{2}+2x-4 }=2\\\\(\sqrt{3x^{2}+2x-4 })^{2} =2^{2}\\\\3x^{2}+2x-4=4\\\\3x^{2}+2x-8=0$

$D=2^{2}-4*3*(-8)=4+96=100=10^{2}\\\\x_{3}=\frac{-2-10}{6}=-2\\\\x_{4}=\frac{-2+10}{6}=1\frac{1}{3}\\\\ODZ:3x^{2} +2x-2\geq0\\\\3x^{2}+2x-2=0\\\\D=2^{2}-4*3*(-2)=4+24=28=(2\sqrt{7})^{2}\\\\x_{1} =\frac{-2-2\sqrt{7} }{6}=\frac{-1-\sqrt{7} }{3}\\\\x_{2}=\frac{-2+2\sqrt{7} }{6}=\frac{\sqrt{7}-1 }{3}\\\\x\in(-\infty \ ; \ \frac{-1-\sqrt{7} }{3}] \ \cup \ [\frac{\sqrt{7}-1 }{3} \ ; \ +\infty)$

Подходят все 4 корня