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### Course: Algebra 2 > Unit 10

Lesson 6: Solving equations by graphing- Solving equations by graphing
- Solving equations by graphing: intro
- Solving equations graphically: intro
- Solving equations by graphing: graphing calculator
- Solving equations graphically: graphing calculator
- Solving equations by graphing: word problems
- Solving equations graphically: word problems
- Equations: FAQ

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# Solving equations by graphing: intro

Some equations are hard to solve exactly with algebraic tools. We can always solve an equation by graphing it, although the solution might not be exact. This is an example of how to solve a relatively simple equation graphically.

## Want to join the conversation?

- When are we allowed to do this? And wouldn't the first example make it a non-invertible function?(3 votes)
- I am pretty sure you are allowed to do this with any equation. I may be wrong, but that is what I think.(6 votes)

- For example 1, which is y=(3/2)^x=5, couldn't you also find the point of intersection with logarithms? (3/2)^x=5 is an exponential expression. Therefore, it could also be written as log_3/2(5)=x. Using the change of base formula, the logarithm could be rewritten as log(5)/log(3/2). When I plug this into my calculator, I get x=3.969362296. I rounded up to 4 to get the answer. I know this example was used to show the usefulness of graphing when approximating an answer; however, I believe this example would also be an opportune time to utilize logarithms, especially if graph paper is not available. If you see any errors in my work or logic, please let me know.(2 votes)
- How do you solve this without using graphs or computers?(1 vote)
- There are generally multiple ways to solve such problems and the possibilities depend on the particular problem.

For the first problem, (3/2)^x = 5, for example, you could find an upper and lower bound for the value of x and then keep shrinking the range of values to get better approximations for x. Alternatively, a more complex solution would involve using some algebra and something called a Maclaurin series (a topic covered in calculus) to directly compute the value for x.

For the second problem, y = x^3 - 2x^2 - x + 1, you could factor the polynomial and see how many distinct solutions there are for each question. You could also compute the derivative (another calculus topic), find the regions over which the function increases/decreases, and check whether each region goes through the particular y-value of each question.

These are only a couple of ways I thought to solve them off the top of my head. There are no doubt other ways to solve them as well.(2 votes)

- I dont understand, for the first question why not just set up a log equation and solve. Are we supposed to be trying to use the graphs is that why he avoids it in this video?(1 vote)
- The lesson is explicited stted as "Solving equations by graphing." So, yes, we intend to solve using only graphs.(1 vote)

## Video transcript

- [Instructor] We're
told, this is the graph of y is equal to 3/2 to the x. And that's it right over there. Use the graph to find
an approximate solution to 3/2 to the x is equal to five. So pause this video and
try to do this on your own before we work on this together. All right, now let's work on this. So they already give us a
hint of how to solve it. They have the graph of y
is equal to 3/2 to the x. They graph it right over here. And this gives us a hint,
and especially because it's, they want us to find an
approximate solution, that maybe we can solve this equation or approximate a solution to
this equation through graphing. And the way we could do
that is we could take each side of this equation
and set them up as a function. We could set y equals to each side of it. So if we set y equals
to the left-hand side, we get y is equal to 3/2 to the x power, which is what they originally give us, the graph of that. And if we set y equal
to the right-hand side, we get y is equal to five. And we can graph that. And what's interesting
here is if we can find the x-value that gives us the same y-value on both of these equations, well that means that those
graphs are going to intersect. And if I'm getting the same
y-value for that x-value in both of these, well then that means that 3/2 to the x is going to be equal to five. And so we could look
at where they intersect and get an approximate sense
of what x-value that is. And we can see it, at least over here, it looks like x is roughly equal to four. So x is approximately equal to four. And if we wanted to, and
we'd be done at that point. If you wanted to, you
could try to test it out. You could say, "Hey does
that actually work out? "3/2 to the fourth power,
is that equal to five?" Let's see, three to the fourth is 81. Two to the fourth is 16. It gets us, it gets us
pretty close to five. 16 times five is 80. So it's not exact, but
it gets us pretty close. And if you had a graphing calculator that could really zoom in
and zoom in and zoom in, you would get a value,
you would see that x is slightly different than x equals four. But let's do another example. The key here is that we
can approximate solutions to equations through graphing. So here we are told, this is the graph of y is equal to, so we have
this third-degree polynomial right over here. Use the graph to answer
the following questions. How many solutions does the equation x to the third minus two
x squared minus x plus one equals negative one have? Pause this video and
try to think about that. Well when we think
about solutions to this, we could say, all right, well
let's imagine two functions, one is y is equal to x to
the third minus two x squared minus x plus one, which we
already have graphed here. And let's say that the other equation or the other function is y
is equal to negative one. And then how many times
do these intersect? That would tell us how
many solutions we have. So that is y is equal to negative one. And so every time they intersect, that means we have a solution
to our original equation. And they intersect one,
two, and three times. So this has three solutions. What about the second situation? How many solutions does the
equation all of this business equal two have? Well same drill, we could set y equals to x to the third minus two x
squared minus x plus one. And then we could think
about another function, what if y is equal to two? Well y equals two would be up over there, y equals two. And we could see it
only intersects y equals all of this business once. So this is only going
to have one solution. So the key here, and
I'll just write it out, and these are screenshots
from the exercise on Khan Academy where
you'd have to type in one, or in the previous example,
you would type in four. But these are examples where
you can take an equation of one variable, set both sides of them independently equal to y, graph them, and then think about where they intersect. Because the x-values where they intersect will be solutions to
your original equation. And a graph is a useful
way of approximating what a solution will be,
especially if you have a graphing calculator or
Desmos or something like that.