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

  • Thread starter Thread starter unique_pavadrin
  • Start date Start date
  • Tags Tags
    Function
AI Thread Summary
The equation |x + a| = |x - b| has a unique solution at x=2, leading to the conclusion that a and b must satisfy the condition 4 + a = b. The initial approach to solving the equation involved equating the expressions directly, but this method was deemed incorrect as it led to a contradiction. A correct interpretation suggests that for the equation to have one solution, the values of a and b must be related such that a = -b is not applicable since x is fixed at 2. The discussion highlights confusion over the proper steps to derive the values of a and b, indicating a need for further clarification on the problem.
unique_pavadrin
Messages
100
Reaction score
0

Homework Statement


The equation \left| {x + a} \right| = \left| {x - b} \right| 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:
<br /> \begin{array}{l}<br /> \left| {x + a} \right| = \left| {x - b} \right| \\ <br /> x + a = x + b \\ <br /> \end{array}<br />
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;
<br /> \begin{array}{l}<br /> x + a = - x + b \\ <br /> 2x + a = b \\ <br /> 4 + a = b \\ <br /> a = k \\ <br /> a = k + 4 \\ <br /> \end{array}<br />

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
 
Physics news on Phys.org
unique_pavadrin said:

Homework Statement


The equation \left| {x + a} \right| = \left| {x - b} \right| 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:
<br /> \begin{array}{l}<br /> \left| {x + a} \right| = \left| {x - b} \right| \\ <br /> x + a = x + b \\ <br /> \end{array}<br />
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;
<br /> \begin{array}{l}<br /> x + a = - x + b \\ <br /> 2x + a = b \\ <br /> 4 + a = b \\ <br /> a = k \\ <br /> a = k + 4 \\ <br /> \end{array}<br />

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:
<br /> \begin{array}{l}<br /> \left| {x + a} \right| = \left| {x - b} \right| \\ <br /> x + a = x + b \\ <br /> \end{array}<br />
If you are taking both to be positive, it should be x + a = x - b
Then you can kill the x's. so a=-b.
 
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
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top