# How do you solve  |x – 4| > |3x – 1|?

Nov 23, 2017

Given:

|x – 4| > |3x – 1|

Assuming that $x \in \mathbb{R}$, the following is an alternate form:

sqrt((x – 4)^2) > sqrt((3x – 1)^2)

Asserting the same inequality on the squares within the radicals:

(x – 4)^2 > (3x – 1)^2

Expand the squares:

x^2 – 8x+16 > 9x^2 – 6x+1

Combine like terms:

$- 8 {x}^{2} - 2 x + 15 > 0$

When we multiply both sides by -1, we must change the direction of the inequality:

$8 {x}^{2} + 2 x - 15 < 0$

We know that the above quadratic will be less than 0 between the roots, therefore, we shall find the roots:

$x = \frac{- 2 \pm \sqrt{{2}^{2} - 4 \left(8\right) \left(- 15\right)}}{2 \left(8\right)}$

$x = \frac{- 2 \pm \sqrt{484}}{16}$

$x = \frac{- 2 \pm 22}{16}$

$x = - \frac{24}{16}$ and $x = \frac{20}{16}$

$x = - \frac{3}{2}$ and $x = \frac{5}{4}$

The inequality is true between these numbers:

$- \frac{3}{2} < x < \frac{5}{4}$