Solve the inequality |3x+7| < |x−4|

Full worked solution

1. Eliminate absolute value signs by squaring

   For any real number $a$, $|a{|}^2=a^2$. Both sides of the given inequality are non-negative, so we can square both sides without reversing the inequality direction:

   $$(3x + 7)^2 < (x - 4)^2$$

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

   $$9x^2 + 42x + 49 < x^2 - 8x + 16$$

   Subtract the right-hand side from both sides:

   $$8x^2 + 50x + 33 < 0$$

   This is the simplified non-modular quadratic inequality to solve.

   Note.

   • Squaring only works when both sides are non-negative, which holds for absolute values.
   • Always rearrange to make the quadratic coefficient positive to simplify interval checking later.

2. Find critical values by solving the quadratic equation

   First solve the corresponding quadratic equation $8x^2+50x+33=0$ using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, where $a=8$, $b=50$, $c=33$. Calculate the discriminant first:

   $$\Delta = b^2 - 4ac = 50^2 - 4(8)(33) = 2500 - 1056 = 1444 = 38^2$$

   Substitute back into the quadratic formula:

   $$x = \frac{-50 \pm 38}{2(8)} = \frac{-50 \pm 38}{16}$$

   Compute the two roots:

   • First root: $x=\frac{-50+38}{16}=\frac{-12}{16}=-\frac{3}{4}$
   • Second root: $x=\frac{-50-38}{16}=\frac{-88}{16}=-\frac{11}{2}$ These critical values are $x=-\frac{11}{2}$ and $x=-\frac{3}{4}$.
   Note.

   • The discriminant here is a perfect square, so the roots are rational, which simplifies the final interval.
   • Double-check expansion of the squared terms to avoid sign errors when rearranging.

3. Identify the solution interval

   The quadratic $8x^2+50x+33$ has a positive leading coefficient, so its graph is an upward-opening parabola. An upward-opening parabola is less than 0 between its two roots. To confirm, we can sketch the left- and right-hand side absolute value functions to see where $|3x+7|$ lies below $|x-4|$:

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