Trying to solve |2x-1|-|x+5|-3=0

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To solve the equation |2x-1|-|x+5|-3=0, three cases were considered based on the value of x. For x > 1/2, the solution is x = 9, and for -5 < x < 1/2, the solution is x = -7/3. However, for x < -5, the derived solution x = 3 is incorrect because it contradicts the assumption that x is less than -5, indicating no roots exist in that interval. A graphical approach is suggested to visualize the solution and confirm the absence of roots for x < -5. The discussion emphasizes checking conditions for the absolute values to ensure all cases are properly evaluated.
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Homework Statement


trying to solve |2x-1|-|x+5|-3=0:

Homework Equations


3. The Attempt at a Solution [/B]
for x>1/2 I got x=9

for -5<x<1/2 I got x=-7/3

but for x<-5 I got -(2x-1)-(-(x+5))-3=0 => x=3 (which is incorrect)

why x=3 is incorrect , or why I shouldn't considerate the case x<-5 ?
 
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bonildo said:

Homework Statement


trying to solve |2x-1|-|x+5|-3=0:

Homework Equations


3. The Attempt at a Solution [/B]
for x>1/2 I got x=9

for -5<x<1/2 I got x=-7/3

but for x<-5 I got -(2x-1)-(-(x+5))-3=0 => x=3 (which is incorrect)

why x=3 is incorrect , or why I shouldn't considerate the case x<-5 ?
Because for that interval, (-∞, -5), you are assuming that x < -5. Therefore x cannot equal +3.
 
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bonildo said:

Homework Statement


trying to solve |2x-1|-|x+5|-3=0:

Homework Equations


3. The Attempt at a Solution [/B]
for x>1/2 I got x=9

for -5<x<1/2 I got x=-7/3

but for x<-5 I got -(2x-1)-(-(x+5))-3=0 => x=3 (which is incorrect)

why x=3 is incorrect , or why I shouldn't considerate the case x<-5 ?

If your assumption of x <= -5 leads to the conclusion that the root is at x = +3 (outside the starting interval), that means that there IS NO ROOT in the interval (-∞,-5). Plot a graph to see why this is so.
 
You should try checking conditions this way although not as efficient of choices:

2x-1 and x+5 both positive;
2x-1 and x+5 both negative;
2x-1 positive and x+5 negative;
2x-1 negative and x+5 positive.

Solve the equation in each of the four conditions and check each result.
 
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