Solve the inequality |5−x| ≤ |2−3x|

Full worked solution

1. Remove absolute values by squaring both sides

   For any real numbers $a$ and $b$, $|a|\le |b|$ is equivalent to $a^2\le b^2$, as both sides are non-negative and squaring is an increasing function for non-negative inputs. Apply this to the given inequality:

   $$(5 - x)^2 \le (2 - 3x)^2$$

   Expand both sides:

   $$25 - 10x + x^2 \le 4 - 12x + 9x^2$$

   Note.

   You could also split into cases based on the sign of the expressions inside each absolute value, but squaring eliminates the need for case analysis.

2. Rearrange into standard quadratic inequality form

   Bring all terms to one side to get a quadratic expression greater than or equal to zero:

   $$0 \le 4 - 12x + 9x^2 - 25 + 10x - x^2$$

   Simplify like terms:

   $$0 \le 8x^2 - 2x - 21$$

   Or equivalently:

   $$8x^2 - 2x - 21 \ge 0$$

   Note.

   Double-check signs when moving terms between sides of the inequality — it is easy to flip a sign accidentally here.

3. Find critical values by solving the quadratic equation

   Solve $8x^2-2x-21=0$ using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, where $a=8$, $b=-2$, $c=-21$:

   $$x = \frac{2 \pm \sqrt{(-2)^2 - 4(8)(-21)}}{2(8)} = \frac{2 \pm \sqrt{4 + 672}}{16} = \frac{2 \pm \sqrt{676}}{16} = \frac{2 \pm 26}{16}$$

   The two critical values are:

   $$x = \frac{2 + 26}{16} = \frac{28}{16} = \frac{7}{4}, \quad x = \frac{2 - 26}{16} = \frac{-24}{16} = -\frac{3}{2}$$

   Note.

   For a quadratic $a{x}^2+bx+c$ with $a>0$, the inequality $a{x}^2+bx+c\ge 0$ holds outside the interval between the two roots.

4. State the final solution set

   Since the quadratic $8x^2-2x-21$ has a positive leading coefficient, it opens upwards, so it is non-negative when $x$ is less than or equal to the smaller root, or greater than or equal to the larger root. The solution is:

   $$x \le -\frac{3}{2} \quad \text{or} \quad x \ge \frac{7}{4}$$

   Note.

   You can test values inside and outside the interval to confirm: $x=0$ gives $|5|\le |2|$, which is false, confirming no solution between $-\frac{3}{2}$ and $\frac{7}{4}$.

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