9514 1404 393
Answer:
x = {-2/5, 2 2/3}
Step-by-step explanation:
There are several ways you can solve this. Maybe the simplest is to take advantage of the relation ...
[tex]\sqrt{x^2}=|x|[/tex]
Squaring the absolute value expression on both sides gives ...
(4x -3)^2 = (x +5)^2
15x^2 -34x -16 = 0 . . . . rewrite to standard form
(5x +2)(3x -8) = 0 . . . . . factor
x = -2/5, +8/3
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Check
To determine if one of these solutions is extraneous, we can try them both in the original equation. We find they both satisfy the equation, so neither found solution is extraneous.
For x=-2/5, |4(-2/5)-3| = |-2/5 +5| ⇒ |-4 3/5| = |4 3/5| . . . . true
For x = 8/3, |4(8/3) -3| = |8/3 +5| ⇒ |(32 -9)/3| = |(8 +15)/3| . . . . true
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Additional comment
Often, we solve absolute value equations using piecewise functions:
|4x -3| = {4x -3, for x ≥ 3/4; -4x +3, for x < 3/4}
|x +5| = {x +5, for x ≥ -5; -x -5, for x < -5}
Now, we have three regions in which to look for solutions: (-∞, -5], [-5, 3/4], [3/4, ∞).
In the region (-∞, -5], the equation is ...
-4x +3 = -x -5 ⇒ x = 8/3 . . . . . no solution in the region
In the region [-5, 3/4], the equation is ...
-4x +3 = x +5 ⇒ x = -2/5 . . . . . a solution in the region
In the region [3/4, ∞), the equation is ...
4x +3 = x +5 ⇒ x = 8/3 . . . . . . a solution in the region
Note that this last equation is identical to the equation for the region (-∞, -5] with both sides multiplied by -1.
Effectively, we have found the two solutions to ±(4x -3) = (x +5). The trick is to figure whether either of those solutions lies in the region for which the respective equation is defined.
