Question

Решите уравнение!!!плз

Answer 1

$\sqrt{3x+4}= 2x-x$
$\sqrt{3x+4} =x$
esli x<0 netu  reshenie
esli x$\geq$0⇒$\left \{ {{3x+4 \geq 0} \atop {3x+47= x^{2} }} \right.$
$\left \{ {{3x \geq -4} \atop { x^{2} -3x-4=0}} \right.$
$\left \{ {{x \geq -4/3} \atop {x=4, x=-1}} \right.$⇒4,-1
$\left \{ {{x \geq 0} \atop {x=4, x=-1}} \right.$⇒x=4

Answer 2

$\sqrt{3x+4} +x=2x \\ \sqrt{3x+4} =2x-x \\ \sqrt{3x+4} =x \\ \\ 3x+4= x^{2} \\\\ - x^{2} +3x+4=0 \\ x^{2} -3x-4=0 \\ D=9+16=25 \\ \sqrt{D} =5 \\ x_{1} = \frac{3+5}{2} = \frac{8}{2} =4 \\ x_{2} = \frac{3-5}{2} =-1$
$3x+4 \geq 0 \\ 3x \geq -4 \\ x \geq -1 \frac{1}{3} \\ \\ x=4;x=-1\ \ (D)$