Main content
Algebra II (2018 edition)
Course: Algebra II (2018 edition) > Unit 5
Lesson 1: Solving square-root equations- Intro to square-root equations & extraneous solutions
- 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
© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice
Solving square-root equations: two solutions
Sal solves the equation 6+3w=√(2w+12)+2w which has two solutions.
Want to join the conversation?
Why are both solutions to this equation non-extraneous? I thought squaring a radical provides two answers--one for if the radical is positive, and one for if it is negative. Thus, any equation we solve by squaring the radical will provide us with two answers, one of which will always be extraneous. Why then do both of the solutions to this equation fit with the positive square root, while only one (-6) fits with the negative?(28 votes)
Nicolas, hi!
If you think about that, there are two potential solutions to the equation aˆ2=bˆ2 (which is an equation you get when solving radical equations: you square both sides):
1) a=b (including the case a=0 and b=0)
2) a=-b (which is the same as -a=b)
If there is a solution for -a=b, it is going to be extraneous to the original equation a=b. However, in this case we have 2 different values of w that satisfy our a=b case: -4 and -6.
The two solutions for a=b case result from the fact that after squaring we get a quadratic equation with 2 solutions for the case a=b. However, it does not necessarily have to be 2 solutions, it can be one or more than 2, including extraneous. It all depends on which equation we get after squaring.
Hope this helps.
Explanation for the exercises here may help you to understand this better (includes cases where we do have extraneous solutions): https://www.khanacademy.org/math/algebra2/radical-equations-and-functions/analyzing-extraneous-solutions-of-square-root-equations/a/analyzing-extraneous-solutions-of-square-root-equations
Have fun:)(20 votes)
at1:44how did 12w came in the equation
w.sq+12w+36
can u plz answer(8 votes)
( 6 + w ) ^2 = ( 6 + w ) ( 6 + w )
Then use FOIL to multiply the two binomials:
First terms : 6*6 = 36
Inner terms : +6w
Outer terms : + 6w
Last terms : w*w = w^2(10 votes)
Can it be determined in advance if an equation will have extraneous solutions, without actually having to plug in the answers in the original equation afterwards?(6 votes)
Hello Hodorius,
This is technically true. You can determine which possible solutions would be extraneous if they were indeed solutions.
This sounds complicated, but it's really just a matter of saying, "if solution x is true, will it be extraneous?"
I won't explain how to go all the way through it, as it is somewhat difficult to explain, but it basically is identifying which possible solutions would make the denominator equal to zero.
As for identifying whether an expression will have one of it's solutions guaranteed to be extraneous, I do not know of one.
For now, the most efficient way is to just do as Sal did above; solve the equation and check them against your original equations.
Hope This Helped!(4 votes)
What property allows Sal to go from a quadratic equation of (w+4)(w+6) at2:58; and simply isolate each part of the equation, (w+4=0 and w+6=0) instead of multiplying them together?(3 votes)
This is the zero product property.
It states that if a and b are two real numbers such that ab=0 then a=0 or b=0 (note that both could equal zero in this mathematical usage of the word "or").
The proof is fairly straightforward. Suppose a and b are two numbers whose product is zero. a can either be equal to zero or not equal to zero. If a is equal to zero then we have shown that either a or b is zero (in this case a, it doesn't matter what b is). But if a is not equal to zero, we can take the equation ab=0 and divide both sides by a to get b=0/a=0.
Another way to say this is that if the product of any two numbers is zero, at least one of those numbers MUST be zero. Another way to think of this is that the product of any two non-zero numbers is NEVER zero.
thus in the video, if (w+4) [a number] and (w+6) [another number] multiply to equal zero, then one of them MUST be zero for the equation to be true. thus he writes that either W+4 = 0 or, if that isn't the case, then w+6 must be 0 for the equation to be true.(6 votes)
- At1:40why did Sal write w2+12w+36?(4 votes)
- Sal did the multiplication for ( 6 + w ) ^ 2.
He put the result into standard form for a polynomial which meant that the terms were arranged from highest exponent to lowest.(1 vote)
What about radical equations with a radical on both sides?(1 vote)
If two radicals are equal, then the radicands are also equal:
√𝑎 = √𝑏 ⇒ 𝑎 = 𝑏
Just remember that 𝑎, 𝑏 ≥ 0(6 votes)
√(2x+9) - √(x+1)=√(x+4)
What are the steps to solving for x?
Do you square both sides or what?(3 votes)
That's a good start. Squaring both sides will leave you with just one radical. Then you can isolate the radical on one side and square again to get rid of it.(3 votes)
Can anyone help me solve this problem with an explanation? It has two solutions.
(7-(2/x))=√(7/x)(3 votes)
The problem you posted has no solution.
(7-2)/x=sqrt(7)/x
5/x=sqrt(7)/x
cross multiplying gives us: 5x=x*sqrt(7)
squaring both sides: 25x^2=7x^2
18x^2=0
x=0
Looking at the problem in question, you can see that x=0 makes it undefined.(2 votes)
Why did you make these equal to 0? In other equations, ie, one solution and no solution, you isolated the radical on one side. Is there an easy way to tell if an equation has two solutions? Is it because there are radicals on both sides of the equal sign?(3 votes)- you make them equal to 0 once you combined the like terms on both sides. once you take one side (in this video 2w+12) and transfer it over to the other side there is nothing left on the side so it just becomes 0.(3 votes)
So here is my example of a square root equation
so the square root of (9-6) cubed - (4-2)
1st- u solve 9-6 which is 3 then you solve 4-2 which is 2 then u subtract 3 cubed (27) by 2 which is 25 then you solve the square root of 25 which is 5. So did i do it right?(3 votes)
1:44Video transcript
- [Voiceover] Let's say
that we have the equation six plus three w is equal to the square root of two w plus 12 plus two w. See if you can pause the
video and solve for w, and it might have more than one solution, so keep that in mind. Alright, now let's work
through this together. So the first thing I'd like to do whenever I see one of
these radical equations is just isolate the radical
on one side of the equation. So let's subtract two w from both sides. I want to get rid of that two
w from the right-hand side. I just want the radical sign. And if I subtract two w from both sides, what am I left with? Well, on the left-hand
side, I am left with six plus three w minus two w. Well, three of something
take away two of 'em, you're going to be left with w. Six plus w is equal to, these cancel out, we're left with the square root of two w plus 12. And to get rid of the radical, we're going to square both sides, and we've seen before that
this process right over here, it's a little bit tricky, because when you're squaring a radical in a radical equation like this, and then you solve, you might
find an extraneous solution. What do I mean by that? Well, we're going to get the same result whether we square this or
whether we square that, because when you square a
negative, it becomes a positive. But those are fundamentally
two different equations. We only want the solutions that satisfy the one that doesn't
have the negative there. So that's why we're going
to test our solutions to make sure they're valid
for our original equation. So, if we square both sides, on the left-hand side,
we're going to have, well, it's gonna be w squared plus two times their product. So, two times six times w, so that's 12 w, plus six squared, 36, Is equal to, now, if
you take the square root and square it, you're going to
be left with two plus twelve. Now, we can subtract two
w and 12 from both sides, so let's do that. So then we can get it into
a standard quadratic form. So let's subtract two w from both sides and let's subtract 12 from both sides. So, subtract 12 from the right. Subtract 12 here. And once again, I just
want to get rid of this on the right-hand side, and I'm going to be left with, I am going to be left with, on the left-hand side, it's gonna be w squared. See, 12 w minus two w is plus ten w, and then 36 minus twelve is plus 24, is equal to zero. Now, let's see, to solve this, let's see, is this factorable? Are there two numbers that up to ten and whose product is 24? What jumps out at me is six and four. So we can rewrite this as w plus four times w plus six is equal to zero. And so, if I have the
product of two things equaling zero, well to solve this,
either one or both of them could be equal zero. Zero times anything is gonna be zero. So, w plus four is equal to zero, or w plus six is equal to zero. And over here, if you
subtract four from both sides, you get w is equal to negative four. Or, subtract six from both sides here. w is equal to negative six. Now let's verify that these
actually are solutions to our original equation. Remember, our original equation was six. I'll rewrite it here. Our original equation was six plus three w is equal to the square
root of two w plus 12 plus two w. So let's see, if w is
equal to negative four, if w is equal to negative four right over, let me do something different, is equal to negative four. So, that's going to be six
plus three times negative four is equal to the square root
of two times negative four plus 12 plus two times negative four. So this would be, this
is, negative twelve here. This is negative eight here. This is negative eight here, so you have six plus negative
twelve, which is negative six, is equal to the square root
of negative eight plus 12, is four, plus negative eight. So that would be negative six is equal to two plus negative eight, which is absolutely true. So this is definitely a solution. And let's try w is equal to negative six. So, we is equal to negative six. So, we're gonna get, if we look
up here, we're going to have six plus three times negative six is equal to the square root
of two times negative six plus 12 plus two w. So, this is gonna be negative 18. This is going to be negative 12. This is negative 12. Negative 12 plus 12 is zero. Square root of zero, this is always zero. And then two times, and actually let me, I shouldn't
have written a w there, I should have written
two times negative six. So back to what I was doing, this right over here is negative eighteen. This is two times negative six plus 12. This is all zero. Square root of zero is zero. And then this is negative twelve. So you get six plus negative of eighteen which is negative 12. Is equal to zero plus negative 12. Negative 12. Which is absolutely correct. So these are actually both solutions to our original radical equation.