Question

find the set of values of x for which:

(a) x^2-3x>=10 and (x-5)^2<4

(b) x^2+4x-21<=0 and x^2-9x+8>0

(c) x^2+x-2>0 and x^2-2x-3>=0

Answer 1

(a)

$$x^2-3x\geq10$$

$$x^2-3x-10\geq0$$

$$(x+2)(x-5)\geq0$$

$$x\in(-\infin, -2]\cup[5, \infin)$$

$$(x-5)^2<4$$

$$-2<x-5<2$$

$$3<x<7$$

Answer: $x\in[5, 7).$

(b)

$$x^2+4x-21\leq0$$

$$(x+7)(x-3)\leq0$$

$$x\in[-7,3]$$

$$x^2-9x+8>0$$

$$(x-1)(x-8)>0$$

$$x\in(-\infin, 1)\cup (8, \infin)$$

Answer: $x\in[-7, 1).$

(c)

$$x^2+x-2>0$$

$$(x+2)(x-1)>0$$

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

$$x^2-2x-3\geq0$$

$$(x+1)(x-3)\geq0$$

$$x\in(-\infin, -1]\cup[3, \infin)$$

Answer: $x\in(-\infin, -2)\cup[3, \infin).$