Solve the inequality |x^2 - 5x + 6| + |x - 4| >= -x^2 + 5x + 7.

Step-by-step solution

1. 1

   Factor inside the absolute value to find every breakpoint. Since $x^2-5x+6=(x-2)(x-3)$, that expression changes sign at $x=2$ and $x=3$; the second absolute value changes at $x=4$. The three breakpoints split the line into four intervals:

   $(-\infty,2),\quad [2,3),\quad [3,4),\quad [4,\infty)$

   On each one, both absolute values become ordinary polynomials and the inequality becomes a quadratic one.

2. 2

   Case $x<2$: both insides positive and negative respectively. Here $(x-2)(x-3)>0$ and $x-4<0$, so the left side is $x^2-5x+6-(x-4)=x^2-6x+10$. The inequality becomes

   $x^2-6x+10\ge -x^2+5x+7\ \Longrightarrow\ 2x^2-11x+3\ge 0$

   Its roots are $x=\dfrac{11\pm\sqrt{97}}{4}$, i.e. about $0.2878$ and $5.212$. An upward parabola is non-negative outside its roots, and intersecting with $x<2$ leaves

   $x\le\frac{11-\sqrt{97}}{4}$

3. 3

   Case $2\le x<3$: the quadratic inside flips sign. Now $(x-2)(x-3)\le 0$, so $|x^2-5x+6|=-x^2+5x-6$, and still $|x-4|=4-x$. The inequality becomes

   $-x^2+4x-2\ge -x^2+5x+7$

   The $-x^2$ cancels on both sides, leaving the linear condition $-9\ge x$. No point of $[2,3)$ satisfies $x\le -9$, so this case is empty.

4. 4

   Case $3\le x<4$: back to a positive quadratic. Here $|x^2-5x+6|=x^2-5x+6$ and $|x-4|=4-x$, giving the same inequality as Case 1, $2x^2-11x+3\ge 0$, whose solution set is $x\le 0.2878$ or $x\ge 5.212$. Neither region meets $[3,4)$, so this case is empty too.

5. 5

   Case $x\ge 4$: both absolute values open positively. Now $|x-4|=x-4$, so

   $x^2-4x+2\ge -x^2+5x+7\ \Longrightarrow\ 2x^2-9x-5\ge 0$

   This factors nicely: $2x^2-9x-5=(2x+1)(x-5)$, with roots $x=-\tfrac12$ and $x=5$. Outside the roots and intersected with $x\ge 4$ this gives

   $x\ge 5$

6. 6

   Union the surviving cases and spot-check.

   $\left(-\infty,\ \frac{11-\sqrt{97}}{4}\right]\cup[5,\infty)$

   Numerically the first endpoint is $0.28779$. Testing: at $x=0$ the left side is $6+4=10$ and the right is $7$, so it holds ✓; at $x=1$ we get $2+3=5$ against $11$, which fails ✓; at $x=5$ both sides equal $7$, so the endpoint is included ✓; at $x=6$ we get $12+2=14$ against $1$ ✓.