Integer inequalities and product of factors

Text

Solve:

• (a) $(3x-2)^2\le (x+3)^2$
• (b) $-x^2-3x+4\le 0$
• (c) $-3x^4+x^3+4x^2\ge 0$
• (d) $(2x+1)(-x-2)(3+2x^2)\le 0$

Solution

(a) Moving everything to the left and using the difference of squares: $(3x-2)^2-(x+3)^2\le 0 \iff \big[(3x-2)-(x+3)\big]\big[(3x-2)+(x+3)\big]\le 0 \iff (2x-5)(4x+1)\le 0.$ The roots are $x=\tfrac52$ and $x=-\tfrac14$; the parabola opens upward, so it is $\le 0$ between the roots: $-\tfrac14\le x\le \tfrac52$.

(b) Changing sign: $-x^2-3x+4\le 0 \iff x^2+3x-4\ge 0 \iff (x+4)(x-1)\ge 0$. Solution outside the roots: $x\le -4\ \vee\ x\ge 1$.

(c) Changing sign and factoring: $-3x^4+x^3+4x^2\ge 0 \iff x^2(-3x^2+x+4)\ge 0$. Since $x^2\ge 0$ (and $=0$ at $x=0$), for $x\neq 0$ the sign depends on $-3x^2+x+4\ge 0 \iff 3x^2-x-4\le 0 \iff (3x-4)(x+1)\le 0$, that is $-1\le x\le \tfrac43$ (an interval that already contains $x=0$). Solution: $-1\le x\le \tfrac43$.

(d) The factor $3+2x^2>0$ always. There remains $(2x+1)(-x-2)\le 0 \iff -(2x+1)(x+2)\le 0 \iff (2x+1)(x+2)\ge 0$. Roots $x=-\tfrac12$ and $x=-2$; solution outside: $x\le -2\ \vee\ x\ge -\tfrac12$.

$\boxed{\begin{aligned} &\text{(a) } -\tfrac14\le x\le \tfrac52; \quad \text{(b) } x\le -4 \vee x\ge 1;\\ &\text{(c) } -1\le x\le \tfrac43; \quad \text{(d) } x\le -2 \vee x\ge -\tfrac12\end{aligned}}$