Answer to Question #232650 in Algebra for dizzo

Question

Solve and give solution set for the following equation in R

(a) 2X+3<3X+4>X+5

(b) 2x^2+4=0

(c) 2sin ( 180-x) =1

(d) (X^2-X-6) ÷ (X^2-X-2)<0

Answer 1

(a)

$$2x+3<3x+4>x+5$$

Then

$$\begin{matrix} 2x+3<3x+4 \\ 3x+4>x+5 \end{matrix}$$

$$\begin{matrix} x>-1 \\ 2x>1 \end{matrix}$$

Answer: $x>\dfrac{1}{2}$

$x\in(\dfrac{1}{2}, \infin)$

(b)

$$2x^2+4=0$$

$$x^2\geq0, x\in\R$$

$$2x^2+4\geq4, x\in\R$$

The given equation has no solution for $x\in\R.$

$$2x^2+4=0$$

$$x^2=-2$$

$$x_1=-i\sqrt{2}, x_2=i\sqrt{2}$$

(c)

$$2\sin(180-x) =1$$

$$2\sin x=1$$

$$\sin x=\dfrac{1}{2}$$

$$x=(-1)^n\cdot30\degree+180\degree n, n\in\Z$$

(d)

$$\dfrac{x^2-x-6}{x^2-x-2}<0$$

$$(x^2-x-6)(x^2-x-2)<0$$

$$(x+2)(x-3)(x+1)(x-2)<0$$

$$x\in(-2, -1)\cup(2, 3)$$