Solve the inequality |4x−3| < |x+6|

Full worked solution

1. (1) Eliminate absolute values by squaring both sides

   For any real number $a$, $|a|^2 = a^2$, and since both sides of the inequality are non-negative, we can square both sides without reversing the inequality direction:

   $$(4x - 3)^2 < (x + 6)^2$$

   Expand both sides and rearrange terms to form a standard quadratic inequality:

   $$16x^2 - 24x + 9 < x^2 + 12x + 36$$

   Subtract the right-hand side from both sides:

   $$15x^2 - 36x - 27 < 0$$

   We can simplify this by dividing all terms by 3 (since 3 is positive, the inequality direction stays the same):

   $$5x^2 - 12x - 9 < 0$$

   Note.

   Only square both sides of an inequality when both sides are non-negative, which is always true for absolute values.

2. (2) Find critical values by solving the quadratic equation

   First solve the corresponding quadratic equation $5x^2 - 12x - 9 = 0$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=5$, $b=-12$, and $c=-9$. Calculate the discriminant first:

   $$\Delta = b^2 - 4ac = (-12)^2 - 4(5)(-9) = 144 + 180 = 324 = 18^2$$

   Substitute back into the quadratic formula:

   $$x = \frac{12 \pm 18}{2(5)} = \frac{12 \pm 18}{10}$$

   Compute the two roots:

   • First root: $x = \frac{12 + 18}{10} = \frac{30}{10} = 3$
   • Second root: $x = \frac{12 - 18}{10} = \frac{-6}{10} = -\frac{3}{5}$

   The critical values are $x = -\frac{3}{5}$ and $x = 3$.

   Note.

   The discriminant tells you the number of real roots: a perfect square here means the quadratic factors cleanly, which we can verify: $5x^2 -12x -9 = (5x + 3)(x - 3)$.

3. (3) Determine the solution interval

   The quadratic $y = 5x^2 - 12x -9$ has a positive leading coefficient, so its graph is an upward-opening parabola. This means the quadratic is less than zero between its two roots, where the graph lies below the x-axis. To confirm, we can test a value between the roots, e.g. $x=0$: $5(0)^2 -12(0) -9 = -9 < 0$, which satisfies the inequality. Testing values outside the interval (e.g. $x=-1$ or $x=4$) gives positive values, which do not satisfy the inequality.

   We include a sketch of the absolute-value functions to confirm the solution:

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