Question

Решите уравнение (х+2)*√(23-4х)-3х^2=0

Answer 1

$(x+2)* \sqrt{(23-4x)-3x^2} =0$
ОДЗ: $23-4x-3x^2 \geq 0$
$3 x^{2} +4x-23 \leq 0$
$D=16+276=292$
$x_1= \frac{-2+ \sqrt{73} }{3}$
$x_2= \frac{-2- \sqrt{73} }{3}$
$x$∈$[ \frac{-2- \sqrt{73} }{3} ; \frac{-2+ \sqrt{73} }{3} ]$
$x+2=0$   или $\sqrt{23-4x-3x^2} =0$
$x=-2$  или $x_1= \frac{-2+ \sqrt{73} }{3}$  $x_2= \frac{-2- \sqrt{73}}{3}$
Ответ: $-2; \frac{-2+ \sqrt{73} }{3} ; \frac{-2- \sqrt{73} }{3}$

Answer 2

$23 -4x -3x^2 \geq 0; \ \ \ 3x^2 +4x-23 \leq 0 \\ \\ 3x^2+4x-23=0; \ \ x_{1,2}=\frac{b^2 \pm \sqrt{b^2 - 4 \cdot a \cdot c}}{2 \cdot a}=\frac{-4 \pm \sqrt{16 +276}}{6}=\frac{-4 \pm 2 \sqrt{73}}{6} \\ \\ x_1= \frac{\sqrt{73}-2}{3}; \ \ x_2=\frac{-2 - \sqrt{73}}{3} \\ \\ x \in [\frac{-2 - \sqrt{73}}{3}}; \frac{\sqrt{73}-2}{3}]$

$x+2 =0; \ \ x=-2 \in [\frac{-2 - \sqrt{73}}{3}}; \frac{\sqrt{73}-2}{3}] \\ \\ OTBET: \ -2; \ \frac{-2 - \sqrt{73}}{3}}, \ \frac{\sqrt{73}-2}{3}$