Question

Solve the following inequality:
$(x^2+3x+1)(x^2+3x-3)⩾5$

Answer 1

$(x^2+3x+1)\left(\right(x^2+3x+3)\ge 5$

Let $x^2+3x$ be t

$(t+1)(t-3)\ge 5t^2-2t-3-5\ge 0t^2-2t-8\ge 0(t-4)(t+2)\ge 0\therefore (x^2+3x-4)\left(\right(x^2+3x+2)\ge 0(x+4\left)\right(x-1\left)\right(x+2\left)\right(x+1)\ge 0x\in (-\infty ,-4)\cup (-2,-1)\cup (1,\infty )$