Find Solutions for |x-1|*|x+1|=0 with Step-by-Step Guide

  • Thread starter solar nebula
  • Start date
In summary, the purpose of finding solutions for |x-1|*|x+1|=0 is to determine the values of x that satisfy the given equation, which can be used to solve other related problems or to graph the equation. The first step in finding solutions is to set the equation equal to 0 and solve for x. To solve for x, you can use the zero product property and set each factor equal to 0. There are restrictions on the values of x, with the solution set being x=1 and x=-1. To graph the solution set, plot the solutions on a number line by drawing vertical lines at x=1 and x=-1. The solution set is where these lines intersect the number line
  • #1
solar nebula
14
0

Homework Statement


find all the numbers x for which |x-1|*|x+1|=0

Homework Equations





The Attempt at a Solution


|x-1|*|x+1|=0

|x-1|=0 or |x+1|=0

x-1=0 or x+1=0

x=1 or x=-1

but I think there's a flaw some in my attempt solution...or the whole thing is a flaw..
any help will be much appreciated!
Thanks!
 
Physics news on Phys.org
  • #2
It's completely correct!
 
  • #3
micromass said:
It's completely correct!

But when I try to do the same thing for |x-1|*|x+2|=3 which is:

|x-1|*|x+2|=3

|x-1|=3 or |x+2|=3

x-1=3 or x+2=3

x=4 or x=1

But this solution does not make sense...
please help, Thanks!
 
Last edited:
  • #4
Of course not. You basically did "xy=3 thus x=3 or y=3". This is obviously false.

To solve that, first do the following:

[tex]|x-1||x+1|=3[/tex]

thus

[tex]|(x-1)(x+1)|=3[/tex]

So the absolute value of the number (x-1)(x+1) equal 3. Then what can the number (x-1)(x+1) equal?

BTW I moved this to precalculus, because it obviously does not belong in calculus and beyond. Please post in the correct forum next time.
 
  • #5
In both cases you are following the procedure
[itex]a*b=c \Rightarrow a=\frac{c}{b}[/itex]
which works as long as [itex]b\neq0[/itex]. When you do it withe [itex]c=0[/itex] then [itex]\frac{c}{b}=0[/itex], but this is true for any [itex]b[/itex] (again, I'm assuming b is not zero). So you just skip right to [itex]a=0[/itex] or [itex]b=0[/itex].

micromass's explanation is probably more intuitive, but I like to think of things algorithmically.
 

FAQ: Find Solutions for |x-1|*|x+1|=0 with Step-by-Step Guide

What is the purpose of finding solutions for |x-1|*|x+1|=0?

The purpose of finding solutions for |x-1|*|x+1|=0 is to determine the values of x that satisfy the given equation. These solutions can be used to solve other related problems or to graph the equation.

What is the first step in finding solutions for |x-1|*|x+1|=0?

The first step is to set the equation equal to 0 and solve for x. This will give the x-values that make the equation true.

How do I solve for x in the equation |x-1|*|x+1|=0?

To solve for x, you can use the zero product property which states that if the product of two factors is equal to 0, then at least one of the factors must be equal to 0. In this case, you can set each factor (|x-1| and |x+1|) equal to 0 and solve for x.

Are there any restrictions on the values of x in the equation |x-1|*|x+1|=0?

Yes, there are restrictions on the values of x. Since the absolute value of any number is always positive, the product of two absolute values can only equal 0 if one or both of the factors is equal to 0. This means that the values of x that make the equation true are x=1 and x=-1.

How do I graph the solution set for |x-1|*|x+1|=0?

To graph the solution set, you can plot the two solutions, x=1 and x=-1, on a number line. Since the absolute value of a number is its distance from 0, you can draw a vertical line at x=1 and x=-1 to represent the two solutions. The solution set for the equation is the points where the vertical lines intersect the number line.

Similar threads

Back
Top