Solve the inequality |3x−5| > 2|x+1|

Full worked solution

1. Eliminate absolute values by squaring both sides

   Since both sides of the inequality are non-negative for all real $x$, we can square both sides without changing the direction of the inequality, using the identity $|a{|}^2=a^2$:

   $$(3x - 5)^2 > \left(2|x + 1|\right)^2$$

   Simplify the right-hand side and expand both sides:

   $$9x^2 - 30x + 25 > 4(x^2 + 2x + 1)$$

   Rearrange all terms to one side to get a standard quadratic inequality:

   $$9x^2 - 30x + 25 > 4x^2 + 8x + 4$$

   $$5x^2 - 38x + 21 > 0$$

   Note.

   Never square an inequality if either side can be negative. Since absolute values are always non-negative, this step is safe here.

2. Find critical values by solving the quadratic equation

   First solve the corresponding quadratic equation $5x^2-38x+21=0$ to find the critical points where the left-hand side equals zero. Use the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ with $a=5$, $b=-38$, $c=21$: Calculate the discriminant first:

   $$\Delta = (-38)^2 - 4(5)(21) = 1444 - 420 = 1024 = 32^2$$

   Substitute back into the quadratic formula:

   $$x = \frac{38 \pm 32}{2(5)} = \frac{38 \pm 32}{10}$$

   The two critical values are:

   $$x = \frac{38 + 32}{10} = \frac{70}{10} = 7, \quad x = \frac{38 - 32}{10} = \frac{6}{10} = \frac{3}{5}$$

   Note.

   The discriminant is a perfect square, so the roots are rational — always check for this to simplify calculations.

3. Determine the solution interval from the quadratic shape

   The quadratic $5x^2-38x+21$ has a positive leading coefficient ($a=5>0$), so its graph is an upward-opening parabola that crosses the x-axis at the critical values $x=\frac{3}{5}$ and $x=7$. For an upward-opening parabola, the expression is greater than zero outside the interval between its roots. We can also test values to confirm:

   • For $x<\frac{3}{5}$ (e.g. $x=0$): $|0-5|=5>2|0+1|=2$, which holds.
   • For $\frac{3}{5}<x<7$ (e.g. $x=1$): $|3-5|=2<2|1+1|=4$, which does not hold.
   • For $x>7$ (e.g. $x=8$): $|24-5|=19>2|8+1|=18$, which holds. Therefore the solution is $x<\frac{3}{5}$ or $x>7$.

   To visualise, we sketch the left-hand side $y=|3x-5|$ and right-hand side $y=2|x+1|$ on the same axes:

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