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### Course: Algebra (all content) > Unit 8

Lesson 2: Solving absolute value equations- Intro to absolute value equations and graphs
- Worked example: absolute value equation with two solutions
- Worked example: absolute value equations with one solution
- Worked example: absolute value equations with no solution
- Solve absolute value equations

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# Intro to absolute value equations and graphs

To solve absolute value equations, find x values that make the expression inside the absolute value positive or negative the constant. To graph absolute value functions, plot two lines for the positive and negative cases that meet at the expression's zero. The graph is v-shaped. Created by Sal Khan and CK-12 Foundation.

## Want to join the conversation?

- How could I work on these functions with two or more absolute values?

like graphing f(x) = |2x-1| -3|x+1|(159 votes)- You'll have to consider 4 cases. In one, 2x+1=0 and -(2x+1)=0 . Do that same for the other absolute value and thus you'll obtain 4 answers, i.e, the line will intersect the x axis at 4 points.(43 votes)

- I don't really understand this. Can anyone help me?(17 votes)
- If it's a negative number that you're trying to find the absolute value of, and there are no other terms attached to it, then the answer is the positive of that number. The same goes for positive numbers, except they stay positive. For example, the absolute value of -2 is 2, and the absolute value of 2 is also 2. I hope this helped!(8 votes)

- At9:01, why is the Y intercept equal to 0? From that point on it makes perfect sense how he would then reach the conclusion that X=-3. I just don't understand why Y=0 in that equation. Mucho appreciation for any help!(9 votes)
- You can't say "y-intercept is equal to 0", it's not correct. That x=-3 you're talking about is the x-intercept. An intercept is the point where the line crosses one of the axes; the x-intercept is the point where it crosses the x-axis. Now, if you take a look at a graph, you'll see that the point where the line crosses the x-axis will always have 0 as its y coordinate!

So if you're looking for the x-intercept, by definition you're looking for the point where y=0. How do you do that? Well, simply by taking your equation, which was y=x+3, and setting y=0. If you solve 0=x+3, and subtract 3 from both sides, you have x=-3.

Hope it helps!(39 votes)

- Could you have an Absolute Value of 0.... If you had |0| it would be 0 but there is no way to have a negative 0 so would you ever write |0|. If so what would it be used for???(6 votes)
- the absolute value for zero is zero because zero is zero spaces away from zero on a number line.(17 votes)

- functions really confuse me so what is a good strategy that will help me understand it better?(12 votes)
- What are you having trouble with? Anything specific?

Well, here's an explanation that will hopefully clear things up.

Say you have a number |x|. Let's say for example, you plug in -1, you would always get that |-1| = 1. But what if you plug in 1? You get |1| = 1. Therefore, if you solve for |x|, it would have both a positive and a negative answer.

So, if you have a very simple equation, 2|x| = 1,

you would get,

|x| = 1/2

x = -1/2 or x = 1/2

The same sort of thing applies when you have an equation like

|x+5| = 1

x+5 = 1 or -1 because of the absolute value sign.

In essence, all you have to know to understand these sorts of equations is to that any term with an absolute value sign around has a single output (a positive number), but can have two different types of inputs for the same output (negative and positive).(19 votes)

- I have the problem |3x-2|=2*sqrt(x+8). The issue is I don't know where to start. Does the absolute value side have more weight and I therefore need to resolve the left side first? Or, does the square root side have more weight so I begin to resolve the right side of the equation? Is it possible to square an absolute value?(12 votes)
- Just like with the simpler problems, start by writing out the two equations you will be solving for (a positive and a negative output), Then solve each equation as you normally would.(10 votes)

- So, the absolute value of any negative or positive number would be positive?(6 votes)
- Yes. Unless you had this for example -|8| which would mean you solve the absolute value which is 8 and multiply it by negative one due to the negative dash. Hope this helped and good luck.(9 votes)

- What is the purpose of an absolute value equation?(4 votes)
- to make our brain hurt(5 votes)

- I know that absolute value is usually used for real numbers, but I want to know what the absolute value of what "i" would be. Since it is imaginary I was wondering if you could even do that. An answer and explanation would be much appreciated.(7 votes)
- |a + bi| = √(a^2 + b^2)

|i| = |0 + i| = √(0^2 + 1^2) = √1 = 1(2 votes)

- .......

i learned this a completely different way.(5 votes)- I've never retained it, because it keeps being taught in ways that have not real world practical application. I don't know why, but I can remember almost everything mathematically most of the time except for certain things like this that we never use. Where is the practical application. There are no word problems on this is in this section. I've been enraged all week because I found out, that I don't remmeber how to solve these equations. It's the why... I need to know why, so I can remember this and what it does and it's all for. I hate being taught shortcuts and how to solve something. I want to see math when I look at things, not solve a formed problem mechanically with not understanding as to how it applies to the world around me. So frustrating... It is not appropriate in my opinion to be taught to skip, right or left side operations to "not change the value of the equal sign." It just makes room for days of mistakes. Horrible. All these new and creative ways of teaching, only create problems when diverting from mathematical standard that should be universal. It is not bx/a is incorrect, because it is too easy to mistake be for a non-coefficent of x when solving for the quadratic formula. Good luck if you write it that way... It took me two days to solve that dumb problem, because right side operations should NEVER be skipped when completing multiple or single squares on nummerous pages of work. That rocket has to lift off and astronaughts certainly did not land on the moon by leaving and room at all for error. Please stop teaching students how to make awesome mistakes. We get that it is hard to land on the moon, but please don't make it more complicated when we are trying to learn how to get there. b/a * x is the correct way to write this. Yes there is a correct way; you may arrive at the same result. We should not rely on intuition or "this is the same thing as this" or this "thing." All these "things" have names. Please review your teaching style.(2 votes)

## Video transcript

Let's do some equations that
deal with absolute values. And just as a bit of a review,
when you take the absolute value of a number. Let's say I take the absolute
value of negative 1. What you're really doing is
you're saying, how far is that number from 0? And in the case of negative 1,
if we draw a number line right there-- that's a very badly
drawn number line. If we draw a number line
right there, that's 0. You have a negative
1 right there. Well, it's 1 away from 0. So the absolute value
of negative 1 is 1. And the absolute value of
1 is also 1 away from 0. It's also equal to 1. So on some level, absolute value
is the distance from 0. But another, I guess simpler way
to think of it, it always results in the positive
version of the number. The absolute value of negative
7,346 is equal to 7,346. So with that in mind, let's try
to solve some equations with absolute values in them. So let's say I have the equation
the absolute value of x minus 5 is equal to 10. And one way you can interpret
this, and I want you to think about this, this is actually
saying that the distance between x and 5 is
equal to 10. So how many numbers that are
exactly 10 away from 5? And you can already think of the
solution to this equation, but I'll show you how to solve
it systematically. Now this is going to be true
in two situations. Either x minus 5 is equal
to positive 10. If this evaluates out to
positive 10, then when you take the absolute value
of it, you're going to get positive 10. Or x minus 5 might evaluate
to negative 10. If x minus 5 evaluated to
negative 10, when you take the absolute value of it, you
would get 10 again. So x minus 5 could also be
equal to negative 10. Both of these would satisfy
this equation. Now, to solve this one,
add 5 to both sides of this equation. You get x is equal to 15. To solve this one, add 5 to both
sides of this equation. x is equal to negative 5. So our solution, there's
two x's that satisfy this equation. x could be 15. 15 minus 5 is 10, take the
absolute value, you're going to get 10, or x could
be negative 5. Negative 5 minus 5
is negative 10. Take the absolute value,
you get 10. And notice, both of these
numbers are exactly 10 away from the number 5. Let's do another one of these. Let's do another one. Let's say we have the absolute
value of x plus 2 is equal to 6. So what does that tell us? That tells us that either x plus
2, that the thing inside the absolute value sign,
is equal to 6. Or the thing inside of the
absolute value sign, the x plus 2, could also
be negative 6. If this whole thing evaluated
to negative 6, you take the absolute value, you'd get 6. So, or x plus 2 could
equal negative 6. And then if you subtract 2
from both sides of this equation, you get x could
be equal to 4. If you subtract 2 from both
sides of this equation, you get x could be equal
to negative 8. So these are the two solutions
to the equation. And just to kind of have it
gel in your mind, that absolute value, you can kind of
view it as a distance, you could rewrite this problem as
the absolute value of x minus negative 2 is equal to 6. And so this is asking me, what
are the x's that are exactly 6 away from negative 2? Remember, up here we said,
what are the x's that are exactly 10 away from
positive 5? Whatever number you're
subtracting from positive 5, these are both 10 away
from positive 5. This is asking, what
is exactly 6 away from negative 2? And it's going to be
4, or negative 8. You could try those numbers
out for yourself. Let's do another one of these. Let's do another one, and
we'll do it in purple. Let's say we have the absolute
value of 4x-- I'm going to change this problem
up a little bit. 4x minus 1. The absolute value of 4x minus
1, is equal to-- actually, I'll just keep it--
is equal to 19. So, just like the last few
problems, 4x minus 1 could be equal to 19. Or 4x minus 1 might evaluate
to negative 19. Because then when you take the
absolute value, you're going to get 19 again. Or 4x minus 1 could be
equal to negative 19. Then you just solve these
two equations. Add 1 to both sides of this
equation-- we could do them simultaneous, even. Add 1 to both sides of this,
you get 4x is equal to 20. Add 1 to both sides of this
equation, you get 4x is equal to negative 18. Divide both sides of this by
4, you get x is equal to 5. Divide both sides of this by
4, you get x is equal to negative 18/4, which is
equal to negative 9/2. So both of these x values
satisfy the equation. Try it out. Negative 9/2 times 4. This will become
a negative 18. Negative 18 minus 1
is negative 19. Take the absolute value,
you get 19. You put a 5 here,
4 times 5 is 20. Minus 1 is positive 19. So you take the absolute
value. Once again, you'll get a 19. Let's try to graph one of
these, just for fun. So let's say I have y is
equal to the absolute value of x plus 3. So this is a function,
or a graph, with an absolute value in it. So let's think about
two scenarios. There's one scenario where the
thing inside of the absolute value is positive. So you have the scenario where
x plus 3-- I'll write it over here-- x plus 3 is
greater than 0. And then you have the scenario
where x plus 3 is less than 0. When x plus 3 is greater than 0,
this graph, or this line-- or I guess we can't call it a
line-- this function, is the same thing as y is equal
to x plus 3. If this thing over here is
greater than 0, then the absolute value sign
is irrelevant. So then this thing is the
same thing as y is equal to x plus 3. But when is x plus
3 greater than 0? Well, if you subtract 3 from
both sides, you get x is greater than negative 3. So when x is greater than
negative 3, this graph is going to look just like y
is equal to x plus 3. Now, when x plus 3
is less than 0. When the situation where this--
the inside of our absolute value sign-- is
negative, in that situation this equation is going to
be y is equal to the negative of x plus 3. How can I say that? Well, look, if this is going to
be a negative number, if x plus 3 is going to be a negative
number-- that's what we're assuming here-- if it's
going to be a negative number, then when you take the absolute
value of a negative number, you're going to
make it positive. That's just like multiplying
it by negative 1. If you know you're taking the
absolute value of a negative number, it's just like
multiplying it by negative 1, because you're going to
make it positive. And this is going to
be the situation. x plus 3 is less than 0. If we subtract 3 from both
sides, when x is less than negative 3. So when x is less than negative
3, the graph will look like this. When x is greater than negative
3, the graph will look like that. So let's see what that
would make the entire graph look like. Let me draw my axes. That's my x-axis, that's
my y-axis. So just let me multiply this
out, just so we have it in mx plus b form. So this is equal to negative
x minus 3. So let's just figure out
what this graph would look like in general. Negative x minus 3. The y-intercept is negative
3, so 1, 2, 3. And negative x means it
slopes downward, has a downward slope of 1. So it would look like this. The x-intercept would be
at x is equal to--. So if you say y is equal to 0,
that would happen when x is equal to negative 3. So it's going to go
through that line, that point right there. And the graph, if we didn't
have this constraint right here, would look something
like this. That's if we didn't constrain
it to a certain interval on the x-axis. Now this graph, what
does it look like? Let's see. It has its y-intercept
at positive 3. Just like that. And where's its x-intecept? When y is equal to 0,
x is negative 3. So it also goes through that
point right there, and it has a slope of 1. So it would look something
like this. That's what this graph
looks like. Now, what we figured out is
that this absolute value function, it looks like this
purple graph when x is less than negative 3. So when x is less than negative
3-- that's x is equal to negative 3 right there-- when
x is less than negative 3, it looks like this
purple graph. Right there. So that's when x is less
than negative 3. But when x is greater than
negative 3, it looks like the green graph. It looks like that. So this graph looks like
this strange v. When x is greater than negative
3, this is positive. So we have the graph of-- we
have a positive slope. But then when x is less than
negative 3, we're essentially taking the negative of the
function, if you want to view it that way, and so we have
this negative slope. So you kind of have this
v-shaped function, this v-shaped graph, which is
indicative of an absolute value function.