How do you solve # |x – 4| |3x – 1|#?

Question

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

Savannah Anderson · Accepted Answer

Given:

#|x – 4| > |3x – 1|#

Assuming that #x in RR#, 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:

#-8x^2-2x+15>0#

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

#8x^2+2x-15<0#

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

#x = (-2+-sqrt(2^2-4(8)(-15)))/(2(8))#

#x = (-2+-sqrt(484))/16#

#x = (-2+-22)/16#

#x = -24/16# and #x = 20/16#

#x = -3/2# and #x = 5/4#

The inequality is true between these numbers:

#-3/2 < x < 5/4#

Genesis Allen · Answer

undefined To solve the inequality |x - 4| > |3x - 1|:

1. Break the inequality into two cases based on the sign of the expression inside the absolute values.

Case 1: When $x - 4 \geq 0$ and $3x - 1 \geq 0$:

- For $x - 4 \geq 0$, solve $x \geq 4$.
   - For $3x - 1 \geq 0$, solve $x \geq \frac{1}{3}$.

Case 2: When $x - 4 \leq 0$ and $3x - 1 \leq 0$:

- For $x - 4 \leq 0$, solve $x \leq 4$.
   - For $3x - 1 \leq 0$, solve $x \leq \frac{1}{3}$.

2. Combine the solutions from both cases.

Therefore, the solution to the inequality is $x \leq \frac{1}{3}$ or $x \geq 4$.

How do you solve # |x – 4| &gt; |3x – 1|#?