What is the solution for the equation |x + a| = |x - b|, given that x=2?

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In summary, the equation |x + a| = |x - b| has one solution at x = 2 and the value of a is equal to -b. However, this solution may not be accurate as it is based on the assumption that x is a fixed point at zero. More information is needed to accurately determine the values of a and b.
  • #1
unique_pavadrin
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Homework Statement


The equation [tex]\left| {x + a} \right| = \left| {x - b} \right|[/tex] has exactly one solution; at x=2. Find the value(s) of a and b

2. The attempt at a solution
Here is the way in which I have approached the situation:
[tex]
\begin{array}{l}
\left| {x + a} \right| = \left| {x - b} \right| \\
x + a = x + b \\
\end{array}
[/tex]
this solution is not possible as the x's on either side of the equation cancel each other out, and a is not equal to b, so;
[tex]
\begin{array}{l}
x + a = - x + b \\
2x + a = b \\
4 + a = b \\
a = k \\
a = k + 4 \\
\end{array}
[/tex]

However I'm not sure on my my final answer, as I have not evaluated the values for a or b, as i have left them in the term of k. However what I have interpreted from the question, I have not been given enough information. Thank you to those who reply, and correct me if anything I have stated is wrong

unique_pavadrin
 
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  • #2
unique_pavadrin said:

Homework Statement


The equation [tex]\left| {x + a} \right| = \left| {x - b} \right|[/tex] has exactly one solution; at x=2. Find the value(s) of a and b

2. The attempt at a solution
Here is the way in which I have approached the situation:
[tex]
\begin{array}{l}
\left| {x + a} \right| = \left| {x - b} \right| \\
x + a = x + b \\
\end{array}
[/tex]
this solution is not possible as the x's on either side of the equation cancel each other out, and a is not equal to b, so;
[tex]
\begin{array}{l}
x + a = - x + b \\
2x + a = b \\
4 + a = b \\
a = k \\
a = k + 4 \\
\end{array}
[/tex]

However I'm not sure on my my final answer, as I have not evaluated the values for a or b, as i have left them in the term of k. However what I have interpreted from the question, I have not been given enough information. Thank you to those who reply, and correct me if anything I have stated is wrong

unique_pavadrin

Not sure how you got this step:
[tex]
\begin{array}{l}
\left| {x + a} \right| = \left| {x - b} \right| \\
x + a = x + b \\
\end{array}
[/tex]
If you are taking both to be positive, it should be [tex]x + a = x - b [/tex]
Then you can kill the x's. so a=-b.
 
  • #3
thanks for the reply theperthvan

ive used those steps as a simply does not equal -b as the x can be anywhere on the number line. by stating that a=-b, x is a fixed point at zero, which is incorrect as z is given the value of 2

or is that completely wrong?
thanks
 

1. What is a piecewise defined function?

A piecewise defined function is a mathematical function that is defined by different rules or formulas for different intervals of its domain. This means that the function behaves differently depending on the value of the input.

2. How do you graph a piecewise defined function?

To graph a piecewise defined function, you need to plot the points for each interval separately and then connect them with a continuous line. It is important to pay attention to the domain and range of each interval in order to accurately plot the function.

3. What is the purpose of using a piecewise defined function?

A piecewise defined function is often used to model real-life situations where different rules or conditions apply to different parts of the problem. It allows for more flexibility and accuracy in representing complex relationships.

4. Can a piecewise defined function be differentiable?

Yes, a piecewise defined function can be differentiable as long as each piece is differentiable and the pieces are connected smoothly at the points where they meet. However, it is important to check the differentiability of each piece separately.

5. How do you determine the domain and range of a piecewise defined function?

The domain and range of a piecewise defined function can be determined by looking at the domain and range of each piece individually and then combining them. It is important to consider any restrictions or exclusions in the domain and range for each piece.

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