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### Course: Algebra 2 > Unit 10

Lesson 2: Square-root equations- Intro to square-root equations & extraneous solutions
- Square-root equations intro
- Intro to solving square-root equations
- Square-root equations intro
- Solving square-root equations
- Solving square-root equations: one solution
- Solving square-root equations: two solutions
- Solving square-root equations: no solution
- Square-root equations

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# Intro to square-root equations & extraneous solutions

Sal explains what square-root equations are, and shows an example of solving such an equation and checking for extraneous solutions. Created by Sal Khan and CK-12 Foundation.

## Want to join the conversation?

- Hello, I have a question/concern . My question/concern is: At 5;53, one can see that the square root of 2.25 is 1.5. However, the square root of 2.25 can also be -1.5. If we put, " -1.5=4.5-6", x being equal to 2.25 would work. So why doesn't x=2.25 work?(29 votes)
- The original equation is (sqrt)x=2x-6. When you see a radical with no + or - sign before it, we assume that we are only taking the principal root (the positive version). When he plugs that 2.25 into the equation, he can't use -1.5 because of that.(27 votes)

- So why are there extraneous solutions?

I know it's because a square root function (parent graph) has only positive y-values. But why can't we take both roots? Why can't there be 2 y-values for every x-value? A square root function is basically a 90 degree rotation of a quadratic function, so why does it only get half?

I know it's because we want it to be a function. But why does it need to be a function? And who decides what a function is anyway? Who says a function can only have one y-value for every x-value?(12 votes)- Interesting question. The thing is a function itself is defined as having one output for every input. If there are two or more outputs for any input, then you no longer have a function. Also, as to "who decides it gets to be a function", pretty much any equation with both variables and constants with non-ambiguous solutions (i.e. y = sqrt(x)) is a function. So let's see how that relates to your claim that a square root function is basically a 90 degree rotation of a quadratic equation.

Sadly I can't put pictures here but if you go to Desmos and graph both`y=sqrt(x)`

and`y=-sqrt(x)`

, you'd see you'd basically get the 90-degree rotated quadratic function. However, now we don't even have a function. There are two values per input. Also, this can't be represented in a single equation/function except for`y=(plus or minus)sqrt(x)`

. But that's really just the two equations we just graphed, since plus or minus is not an operator but rather the choice between addition or subtraction. The plus or minus operator cannot be used for functions because it just splits the whole equation into two (although it could be used to find solutions of a function like in the quadratic formula).

So in short, the square-root equation is a function by nature because it returns one value per input. Keeping it as a function makes everything clearer. Using the non-function definition of square root really gives you two equations instead of one, so for simplicity, the square root in math returns a positive value by nature. Extraneous solutions from radical equations exist whenever it is assumed that the principal root can return two values in one function.(13 votes)

- At1:25or so, Sal multiplies 2x and -6 to get -12x which doubles to -24x. Why? I was understanding everything up till that point.(13 votes)
- The reason why is that when you multiply 2x and -6 you have to do it twice have you ever heard of the F.O.I.L. method

F.O.I.L. is for squaring a set of two numbers it stands for First Outer Inner Last

For example if you do (3x+2)^2 you would rewrite it as (3x+2)(3x+2)

Then you would do 3x^2+2^2+3x*2^2(12 votes)

- At2:48, why did he take 25^2 and not -25^2 like he did with the -b part of the abc formula?(12 votes)
- No, actually Sal can use either 25 squared
**or**-25 squared. I think it is better for Sal to say 25 squared because saying it like that is easier for people.(8 votes)

- What is the definition of "extraneous" and how do I know that an answer is extraneous?(5 votes)
- “Extraneous” means “irrelevant or unrelated to the subject being dealt with”. In math, an extraneous solution is a solution that emerges during the process of solving a problem but is not actually a valid solution. You can only find out whether or not a solution is extraneous by plugging the solution back into the original equation.(17 votes)

- At5:38, Sal uses the "principlal" root of 2.25. Is the principal root always the positive root?(7 votes)
- Yes, the principal root is always the positive root.

If the problem has a minus in front of the square root, then it is asking for the negative root.

For example

- sqrt(2.25) = -1.5 (the minus in front is asking for the negative root).

hope this helps.(8 votes)

- How'd you get 625-4 x 4 x 36/8?(10 votes)
- The quadratic formula is what he used and can be used to solve for x without using factoring or the completing the square method. The formula is this:

x= (-b plus or minus the square root of (b^2 - 4ac)) over 2a.(2 votes)

- what is 4 radical 3(2 radical 6)(8 votes)
- / is radical

4/3*2/6=8/18=8*3/2=24/2

so 24 radical 2(3 votes)

- why does the 24x become 25x after subtracting the x from both sides around2:20?(6 votes)
- This is because after you subtract from a negative, it only adds the digit.

Like -4 - 2 = -6.

Thus -24x - x =-25x.(6 votes)

- At9:15Sal tells that "only 4 satisfies this interpretation".

The interpretation he means is x = (2x-6)^2

Opposing interpretation is x = (6-2x)^2

We have two roots: 4 and 2.25 (extraneous)

I plug both in to**x = (2x-6)^2****4**

4 = (2*4 - 6)^2

4 = (8 - 6)^2

4 = (2)^2

4 = 4*valid***2.25**

2.25 = (2*2.25-6)^2

2.25 = (4.5-6)^2

2.25 = (1.5)^2

2.25 = 2.25*valid*

other interpretation**x = (6-2x)^2****4**

4 = (6-8)^2

4 = (-2)^2

4 = 4*valid***2.25**

x = (6-2x)^2

2.25 = (6-4.5)^2

2.25 = (1.5)^2

2.25 = 2.25*valid*

So what does Sal mean here, if we see that both roots fit?(5 votes)- First, you need to check the solutions using the original equation (it has square roots). The process of squaring both sides of the equation to eliminate the square roots causes sign changes. We use it to help us solve the equation and find potential solutions. But, you can't use it again in the check as it will make the extraneous solution look correct (as you have demonstrated)

Second, you have a sign error: 4.5-6 = -1.5, not 1.5

Sal does the checks correctly (using the original equation and not squaring the equation). When he checks the solution 2.25, the check fails with 1.5 = -1.5

Hope this helps.(6 votes)

## Video transcript

In this video, we're going to
start having some experience solving radical equations or
equations that involve square roots or maybe even higher-power
roots, but we're going to also try to understand
an interesting phenomena that occurs when
we do these equations. Let me show you what
I'm talking about. Let's say I have the equation
the square root of x is equal to 2 times x minus 6. Now, one of the things you're
going to see whenever we do these radical equations is we
want to isolate at least one of the radicals. There's only one of them
in this equation. And when you isolate one of the
radicals on one side of the equation, this one starts
off like that, I have the square root of x isolated on
the left-hand side, then we square both sides
of the equation. So let us square both sides
of the equation. So I'll just rewrite it again. We'll do this one slowly. I'm going to square that and
that's going to be equal to 2x minus 6 squared. And squaring seems like
a valid operation. If that is equal to that, then
that squared should also be equal to that squared. So we just keep on going. So when you take the square root
of x and you square it, that'll just be x. And we get x is equal to-- this
squared is going to be 2x squared, which is 4x squared. It's 2x squared, the
whole thing. 4x squared, and then you
multiply these two, which is negative 12x. And then it's going to be twice
that, so minus 24x. And then negative 6 squared
is plus 36. If you found going from this to
this difficult, you might want to review multiplying
polynomial expressions or multiplying binomials, or
actually, the special case where we square binomials. But the general view, it's this
squared, which is that. And then you have minus 2 times
the product of these 2. The product of those two is
minus 12x or negative 12x. 2 times that is negative 24x,
and then that squared. So this is what our equation
has I guess we could say simplified to, and let's see
what happens if we subtract x from both sides of
this equation. So if you subtract x from both
sides of this equation, the left-hand side becomes zero
and the right-hand side becomes 4x squared minus
25x plus 36. So this radical equation his
simplified to just a standard quadratic equation. And just for simplicity, not
having to worry how to factor it and grouping and all of
that, let's just use the quadratic formula. So the quadratic formula tells
us that our solutions to this, x can be negative b. Negative 25. The negative of negative 25 is
positive 25 plus or minus the square root of 25 squared. 25 squared is 625, minus 4 times
a, which is 4, times c, which is 36, all of that over 2
times 4, all of that over 8. So let's get our calculator
out to figure out what this is over here. So let's say so we have 625
minus-- let's see., this is going to be 16 times 36. 16 times 36 is equal to 49. That's nice. It's a nice perfect square. We know what the square
root of 49 is. It's 7. So let me go back
to the problem. So this in here simplified
to 49. So x is equal to 25 plus or
minus the square root of 49, which is 7, all of
that over 8. So our two solutions here, if we
add 7, we get x is equal to 25 plus 7 is 32, 32/8,
which is equal to 4. And then our other solution,
let me do that in a different color. x is equal to 25 minus
7, which is 18/8. 8 goes into 18 two times,
remainder 2, so this is equal to 2 and 2/8 or 2 and 1/4
or 2.25, just like that. Now, I'm going to show
you an interesting phenomena that occurs. And maybe you might want to
pause it after I show you this conundrum, although I'm going
to tell you why this conundrum pops up. Let's try out to see if our
solutions actually work. So let's try x is equal to 4. If x is equal to 4 works, we
get the principal root of 4 should be equal to 2
times 4 minus 6. The principal root of
4 is positive 2. Positive 2 should be equal to 2
times 4, which is 8 minus 6, which it is. This is true. So 4 works. Now, let's try to do
the same with 2.25. According to this, we should
be able to take the square root, the principal root of
2.2-- let me make my radical a little bit bigger. The principal root of 2.25
should be equal to 2 times 2.25 minus 6. Now, you may or may not be able
to do this in your head. You might know that the square
root of 225 is 15. And then from that, you might
be able to figure out the square root of 2.25 is 1.5. Let me just use the calculator
to verify that for you. So 2.25, take the square root. It's 1.5. The principal root is 1.5. Another square root
is negative 1.5. So it's 1.5. And then, according to this,
this should be equal to 2 times 2.25 is 4.5 minus 6. Now, is this true? This is telling us that 1.5
is equal to negative 1.5. This is not true. 2.5 did not work for this
radical equation. We call this an extraneous
solution. So 2.25 is an extraneous
solution. Now, here's the conundrum: Why
did we get 2.25 as an answer? It looks like we did very valid
things the whole way down, and we got a quadratic,
and we got 2.25. And there's a hint here. When we substitute 2.25,
we get 1.5 is equal to negative 1.5. So there's something here,
something we did gave us this solution that doesn't quite
apply over here. And I'll give you
another hint. Let's try it at this step. If you look at this step, you're
going to see that both solutions actually work. So you could try it
out if you like. Actually, try it out
on your own time. Put in 2.25 for x here. You're going to see
that it works. Put in 4 for x here and you see
that they both work here. So they're both valid
solutions to that. So something happened when we
squared that made the equation a little bit different. There's something slightly
different about this equation than that equation. And the answer is there's two
ways you could think about it. To go back from this equation to
that equation, we take the square root. But to be more particular about
it, we are taking the principal root of both sides. Now, you could take the negative
square root as well. Notice, this is only taking
the principal square root. Going from this right here--
let me be very clear. This statement, we already
established that both of these solutions, both the valid
solution and the extraneous solution to this radical
equation, satisfy this right here. Only the valid one satisfies
the original problem. So let me write the equation
that both of them satisfy. Because this is really an
interesting conundrum. And I think it gives you a
little bit of a nuance and kind of tells you what's
happening when we take principal roots of things. And why when you square both
sides, you are, to some degree, you could either think
of it as losing or gaining some information. Now, this could be written
as x is equal to 2x minus 6 squared. This is one valid interpretation
of this equation right here. But there's a completely other
legitimate interpretation of this equation. This could also be x is
equal to negative 1 times 2x minus 6 squared. And why are these equal
interpretations? Because when you square the
negative 1, the negative 1 will disappear. These are equivalent
statements. And another way of writing
this one, another way of writing this right here, is
that x is equal to-- you multiply negative
1 times that. You get negative 2x plus 6
or 6 minus 2x squared. This and this are two ways
of writing that. Now, when we took our square
root or when we-- I guess there's two ways you
can think about it. When we squared it, we're
assuming that this was the only interpretation, but
this was the other one. So we found two solutions to
this, but only 4 satisfies this interpretation
right here. I hope you get what I'm saying
because we're kind of only taking-- you can imagine the
positive square root. We're not considering the
negative square root of this, because when you take the square
root of both sides to get here, we're only taking
the principal root. Another way to view it--
let me rewrite the original equation. We had the square root of x
is equal to 2x minus 6. Now, we said 4 is a solution. 2.25 isn't a solution. 2.25 would've been a solution if
we said both of the square roots of x is equal
to 2x minus 6. Now you try it out and
2.25 will have a valid solution here. If you take the negative square
root of 2.25, that is equal to 2 times 2.25, so that
is equal to 4.5 minus 6, which is negative 1.5. That is true. The positive version is where
you get x is equal to 4. So that's why we got
two solutions. And if you square this-- maybe
this is an easier way to remember it. If you square this, you actually
get this equation that both solutions are valid. Now, you might have found
that a little bit confusing and all of that. My intention is not
to confuse you. The simple thing to think about
when you are solving radical equations is, look,
isolate radicals, square, keep on solving. You might get more
than one answer. Plug your answers back in. Answers that don't work, they're
extraneous solutions. But most of my explanation in
this video is really why does that extraneous solution
pop up? And hopefully, I gave you some
intuition that our equation is the square root of x. The extraneous solution would
be valid if we took the plus or minus square root of x, not
just the principal root.