Answer to Question #223088 in Algebra for Bawe Bilanyuy

Question

Solve the inequality

x-3 ≤ 2x+5 < -x+13

√(x²-2x-3) ≥ Ix-2I - 2x

Answer 1

1.

$$\begin{cases} x-3\leq2x+5 \\ 2x+5<-x+13 \end{cases}$$

$$\begin{cases} x\geq-8 \\ 3x<8 \end{cases}$$

$$\begin{cases} x\geq-8 \\ x<8/3 \end{cases}$$

Answer: $-8\leq x<8/3$

$x\in[-8, 8/3)$

2.

$$\sqrt{x^2-2x-3} ≥ |x-2|- 2x$$

$$x^2-2x-3\geq0=>(x+1)(x-3)\geq0$$

$$x\leq-1\ or\ x\geq3$$

$x\leq-1$

$$\sqrt{x^2-2x-3} ≥ 2-x-2x$$

$$\sqrt{x^2-2x-3} ≥ 2-3x$$

$$2-3x>0, if\ x\leq-1$$

$$x^2-2x-3\geq(2-3x)^2$$

$$x^2-2x-3\geq4-12x+9x^2$$

$$8x^2-10x+7\leq0$$

$$D=(-10)^2-4(8)(7)=-124<0$$

$$8>0, D<0, 8x^2-10x+7>0, x\in \R$$

There are no solutions for $x\leq-1.$

$x\geq3$

$$\sqrt{x^2-2x-3} ≥x-2-2x$$

$$\sqrt{x^2-2x-3} ≥ -x-2$$

$$-x-2<0, if\ x\geq3$$

Then

$$\sqrt{x^2-2x-3}\geq0>-x-2, x\geq3$$

Answer: $x\geq3$

$x\in[3,\infin)$