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## Algebra II (2018 edition)

### Course: Algebra II (2018 edition) > Unit 5

Lesson 2: Extraneous solutions of radical equations# Equation that has a specific extraneous solution

Sal finds the value of d for which √(3x+25)=d+2x has an extraneous solution at x=-3.

## Want to join the conversation?

- Wouldn't (d+2x)^2 result in (d^2+4xd+4x^2)? I keep trying it at getting the same result.(110 votes)
- You are correct.

If you think you have spotted an error, please report it.(55 votes)

- It's -sqrt(16) the same as sqrt(-16)?(11 votes)
- The square root of 16 times negative 1, would be:
`-sq rt(16) = -(4) = -4`

The square root of -16 itself however, would be 4i.

So they aren't the same; one is -4, the other is 4i. Did that help?(26 votes)

- When asked to pause the vid and try ourselves, I just inputted -3 for x and solved for d, which came back as:

4=d-6 or 10=d

As long as d is not equal to 10 then x=-3 would be an extraneous solution. However, Sal got 2=d, so why did he go for something so specific? What am I missing here?

If someone would explain Sal's reasoning to me, I'd be immensely grateful.(6 votes)- I did it the same way as you initially and after some consideration, I think the issue is that an extraneous solution doesn't simply mean a solution that makes the equation untrue; it is, by definition, "a solution that emerges from solving the problem but is not a valid solution to the problem". The method you used was to simply solve for a d that would prove the equation true, then concluded that any other d would not be true. However, for it to be an extraneous solution, we would have had to come across a value for d through the process of solving the equation that does not actually fit the original equation.

For example, in previous videos Sal demonstrates that there occurs extraneous solutions which solve an equation that was a valid algebraic manipulation of the original equation, but not the original equation itself due to using an operation that cannot be reversible ie taking the square root of both sides.

So in conclusion, an extraneous solution is not simply just any number plucked out that does not prove true the original equation, but a solution that was algebraically extracted through valid operations that does not solve the original equation due to the reversibility of operations used.(8 votes)

- Also, at1:44Sal says, "And so when you solve this purple equation... a quadratic... you'll get two solutions. And it turns out, one of the solutions is going to be for this yellow equation, and one of the solutions is going to be for the purple [red alternate one on the left] equation."

But, in the lesson 'Solving square-root equations: two solutions', we saw a situation where we squared both sides of a radical equation, got a quadratic, ended up with two solutions, and neither solution was extraneous. There is still that alternate equation looming about that may bring about extraneous solutions, since we are squaring both sides of our radical equation, yet neither solution satisfies the alternative extraneous-spawning equation that we only see when we reverse, they both are valid and satisfy the original equation. So, how can he say here that one solution will be for the original equation and one will be for the alternate 'hidden' equation - it seems that this can't be guaranteed, as it's possible for neither one to be for the alternate equation.

Does he just know that it will be the case for this exact problem, but it's not necessarily true in general? Or is it something else about this equation specifically that I'm missing? Because clearly there are situations where one of the solutions is not for the alternate equation, and rather both are for the original... thanks to anyone who can explain :)(8 votes) - Why did he write the negative equation (the one in pink on the left)?(3 votes)
- He was just showing that even if you had a negative square root, it would still work out to the same quadratic, however, the solutions to that quadratic may not work out for the negative square root and the positive one.(6 votes)

- How would squaring two sides keep each side equal? You would be doing different things to each side. For example sqrt(x+3)=5x then you would be multiplying one side by x+3 and the other by 5x(1 vote)
- Remember, the 2 sides are currently equal, but shown using different symbols.

So, you are squaring equal values.

If you have: sqrt(49) = 7

These are currently equal. Both sides are 7, but written in different ways.

Now, square both sides: sqrt(49)^2 = 7^2

You get: 49 = 49

They are still equal.

Hope this helps.(6 votes)

- did anyone else see that Sal made a mistake and said that the square root of (3x+25) = d+2x equals 3x+25=d^2+4dx+x^2 instead of 3x+25=d^2+4dx+4x^2?(2 votes)
- Yeah, but you will see that there is a box that says around1:05, "Sal wrote x² but it should be 4x².", so next time check if there a correction box, if not report the problem (if you are in full screen then it shouldn't appear).(3 votes)

- What is the definition of Radical solution?(2 votes)
- A radical solution is rather simply a solution to a radical equation. There can be none, one or two for any given problem in the unit we study.(1 vote)

- (d+2x)^2= d^2+4dx+4x^2

but he puts it equals d^2+4dx+x^2(2 votes)- I saw the correction box at1:03too. Khan ought to review his videos! :)(1 vote)

- Why does the square root have a negative in front?(2 votes)

## Video transcript

- [Voiceover] We're
asked, which value for D, and we see D in this equation here, makes X equals negative
three an extraneous solution for this radical equation? Squared of 3x plus 25
is equal to D plus 2x. And I encourage you to
pause the video and try to think about it on your own before we work through it together. Alright, now let's work
through this together. So the first thing that
just to remind ourselves is what is an extraneous solution? Well that's a solution that
we get or we think we get but it's really just a
by-product of how we solved it but isn't going to be an actual solution of our original equation. Now how do these extraneous
solutions pop up? Well, it pops up when you
take the square of both sides. So for this equation right
over here, to get rid of the radical, I'd want
to square both sides of it. If I square both sides, the
lefthand side will become 3x plus 25 and the righthand
side, if I square this, is going to be, what? It's going to be D squared
plus 4, 4dx plus X squared. So that's just squaring
both sides of this, but notice, there's actually
a different equation than this one that if you squared
both sides, you would also get this. What is that different equation? Well the different equation
is if you took the negative of one of these sides. So for example, if you had,
if you started the original equation, the negative
square root of 3x plus 25 plus 25, is equal to D plus 2x. You square both sides of this,
you still get this purple equation because you square a negative, you get a positive. So both of them, when
you square both sides, get us over here. And so when you solve
this purple equation, this is a quadratic right over here. You just rearrange it a little bit. You get into standard quadratic form, you'll get two solutions. And it turns out, one of
the solutions is going to be for this yellow equation
and one of the solutions is going to be for the purple equation. And if the solution that
is for the purple equation, is going to be an extraneous solution for the yellow equation. This is actually not
going to be a solution for the yellow equation. So when they say, "Which
value for D makes X equals "negative three and extraneous solution "for this yellow equation?" That's the same thing as saying, "What value of D makes
X equals negative three "a solution for this?" So a solution for this. If it's a solution for this,
it's going to be an extraneous solution for that 'cause these
are two different equations. We're taking the negative of just one side of this equation to get this one. If you took the negative
of both sides of this and that becomes the same
thing 'cause you could multiply both sides of an
equation times a negative value. So a solution for this, which
is equivalently a solution which is equivalent to a
solution, if I, instead of putting the negative on the lefthand
side, if I multiply the righthand side by the negative. But anyway, let's think
about which value for D makes X equals negative
three a solution for this? Well, let's substitute X
equals negative three here and then we just have to solve for D. If X equals negative three,
this is going to be negative the square root of three
times negative three is negative nine, plus 25 is equal to D. Two times negative three is negative six. So D minus 6, and so now we can square both sides of, we can square, actually let's do it this way. We can, I don't wanna
square both sides 'cause we lose some information. It's going to be the negative
square root of negative nine plus 25 is 16, is equal to D minus 6, so this is going to be
equal to negative four. Principle over 16 is four. We have the negative out front, is equal to D minus six and
then add six to both sides. You get two is equal
to, two is equal to D. So if D is equal to two here, if D is equal to two, then
a solution to this purple equation is going to be
X equals negative three. And so that would be an
extraneous solution because if X equals negative three
satisfies this over here, it's definitely gonna
satisfy this over here, but it's not going to
satisfy this up here. And you could verify this if this is equal to two, try out X equals negative three. You're gonna get on the lefthand side, you're going to get 16
and on the righthand side, you're gonna get two minus six which is equal to negative four. Two minus six which is negative four, so this does not work out. X equals negative three
is not a solution to this, but it is a solution for
this and it is a solution to this quadratic right over here. So D equals two makes
X equal negative three, an extraneous solution for this equation.