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## Algebra (all content)

### Course: Algebra (all content) > Unit 12

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

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# Solving square-root equations: no solution

CCSS.Math:

Sal solves the equation √(3x-7)+√(2x-1)=0, only to find out that the single solution is extraneous, which means the equation has no solution.

## Want to join the conversation?

- What's wrong with leaving the terms on one side and squaring everything:

(sqrt(3x-7) + sqrt(2x-1))^2 = 0^2(12 votes)- The radicals will not go away. When you square the left side, you need to use FOIL / distribution to do it. You would get: 3x - 7 + 2*sqrt(3x-7)*sqrt(2x-1) + 2x - 1

Notice, you now have a very complicated radical in the middle.

If you move one square root to the other side, then square the equation, both radicals will get eliminated and you end up with: 3x-7 = 2x-1(17 votes)

- Doesn't this equation have any possible solutions?

Not even a complex solution or something else of the sort?

Can that actually happen that a equation has no possible solution in any number system?(10 votes)- Hi Rutwik, yes an equation can have no possible solution. It helps to think of the equation sqrt(3x-7) = - sqrt(2x-1) as two line graphs, one with the equation y = sqrt(3x-7) and the other y = -sqrt(2x-1) , with both graphs having no common intersection point.

Also, according to the Fundamental Theorem of Algebra, the equation { 3x - 7 = 2x + 1 } has only one root, which only exists if both sides have the same sign. In the initial equation, both sides do not have equivalent signs {sqrt(3x-7) = - sqrt(2x-1)} , which also hints that they do not have a root. Similarly, the equation {-sqrt(3x-7) = sqrt(2x-1)} has no root, while the equations {-sqrt(3x-7) = -sqrt(2x-1)} and {sqrt(3x-7) = sqrt(2x-1)} each have one root.(11 votes)

- At @1:03which is just when we isolate √(3x-7) on the left by subtracting √(2x-1) from both sides of the equation, to have

√(3x-7 =`-`

√(2x-1)

Wouldn't we know right then that there is no solution?

Under what conditions could we have the square root of one expression equal the negative of the square root of the other.(10 votes)- It would still be possible for there to be a solution if both 3x-7 and 2x-1 were equal to 0. Obviously that is NOT the case, but the negative sign on the one radical doesn't eliminate the possibility of a solution by itself.(10 votes)

- Is there a certain rule of : whether the equation has one solution, two solution, or no solution(9 votes)
- Thats exactly what I was wondering. Is there a simple way of knowing if it is 2, 3, or 0 solutions ?(0 votes)

- Ok, so there must have been a way to solve it. My question is that could you have used FOIL and squared both sides of the equation shown above to get the answer?(2 votes)
- It would still be a no solution, but the procces will be much more complicated, because you will have a radical composed of both square roots in the middle.(1 vote)

- It can also be done by using the (a+b)(a-b) = a^2-b^2 formula. You do that then by multiplying (√3x-7-√2x-1) in both sides.

1. (√3x-7+√2x-1) (√3x-7-√2x-1)=0 (√3x-7-√2x-1)

2. (√3x-7)^2 - (√2x-1)^2 = 0

3. (3x-7) - (2x-1) = 0

4. 3x-7-2x+1= 0

5. 3x-2x= 7-1

6. x= 6

Just laying it out there in case if someone thinks that is a better method(4 votes) - at1:42didn't sal make a mistake?

because (-√2x-1)^2=(-(2x-1))^0.5*2=(-2x+1)

but sal wrote (-√2x-1)^2=2x-1(2 votes)- (−√(2𝑥 − 1))^2

= (−(2𝑥 − 1)^0.5)^2

= (−1)^2⋅(2𝑥 − 1)^0.5⋅2

= 1⋅(2𝑥 − 1)

= 2𝑥 − 1(2 votes)

- why not squre the equation without moveing the 2x-1 to the right side?(2 votes)
- You could do that; however, you can't simply square the left side and then remove the square roots. The reasoning why is this:

(a+b)² = a²+2ab+b²

(Let a = sqrt(3x-7) and b = sqrt(2x-1))

This property — more specifically the 2ab — makes squaring the entire left side mathematically labor intensive.

but if you're up w a challenge go for it:) —**however**, if this type of question is in a quiz, Sal's method is more faster.(3 votes)

- What happens if you have two radicals and the term on the other end is not zero. How would you solve that?(4 votes)
- think it's also no solution since you can't even calculate that(0 votes)

- How come he did not just square both sides from the beginning?(2 votes)
- If he had squared both sides at the beginning, the radicals would not be eliminated.

To square`√(3x-7)+√(2x-1)`

you would need to multiply:

(√(3x-7)+√(2x-1)) (√(3x-7)+√(2x-1)) using FOIL or extended distribution. You would end up with:`3x - 7 + 2√[(3x-7)√(2x-1)] + 2x -1`

Hope fully you can see that we still have a square root and the problem just got a lot more complicated than Sal's technique.

Hope this helps.(2 votes)

## Video transcript

- [Voiceover] Let's say that
we have the radical equation, the square root of three x minus seven plus the square root of two x minus one is equal to zero. I encourage you to pause the video and see if you can solve for x before we work through it together. Alright, so one thing we could do is we could try to isolate
each of the radicals on either side of the equation. So let's subtract this one from both sides so I can get it onto the right-hand side, or a version of it on the right-hand side. So I'm subtracting it
from the left-hand side and from the right-hand side. And so, this is going to get us, that is going to get us, on the left-hand side I
just have square root, these cancel out so I'm just left with the square root of three x minus seven is going to be equal to this, the negative of the square
root of two x minus one. So now we can square both sides. And we always have to be
careful when we're doing that because whether we're
squaring the positive or the negative square root
here, we're going to get the same value. So the solution we might
get might be the version when we're solving for
the positive square root, not when we take the negative of it. So we have to test our
solutions at the end to make sure that they are actually valid for our original equation. But if you square both
sides, on the left-hand side we're going to get three x minus seven, and on the right-hand
side, a negative square's just a positive, and the square root of two x minus one squared is
going to be two x minus one. Now let's see, we can
subtract two x from both sides to get all of our xs on one side, so I'm trying to get rid of this. And we can add seven to both sides because I'm trying to get
rid of the negative seven. So add seven to both sides. And we are going to get, we are going to get
three x minus two x is x, is equal to negative one plus seven. x is equal to six. Now let's verify that this actually works. So, if we look at our original equation, the square root of three times six minus seven, minus seven, needs to plus, plus, the square root of
two times six minus one, needs to be equals to zero. So does this actually work out? Three times six minus seven. So this is going to be
the square root of eleven plus the square root of eleven needs to be equal to zero, which clearly is not
going to be equal to zero. This is two square roots of eleven, which does not equal zero. So this does not work, and you might say, wait, how did this happen? I did all of this nice, neat algebra, I didn't make any mistakes, but I got something that doesn't work. Well, this right here is
an extraneous solution. Why is it an extraneous solution? Because it's actually the
solution to the equation, it's a solution to the equation, the square root of three x minus seven minus the square root of two x minus one is equal to zero. And you might say, well,
if it's a solution to that, if it's the solution to
this thing right over here, how did I get the answer while I'm trying to do
algebraic steps there? Well, the key is when we added, when we took this on the
right-hand side and squared it, well it all boiled down to this, regardless of what starting
point you started with, due to the exact same thing
you would've gotten to that say, this point right over here. So the solution to this ended up being the solution to this starting point, versus the one that we
originally started with. So this one, interestingly,
has no solutions. And it'd actually be fun to think about why it has no solutions. We've shown, to a certain degree, the only solution you got by taking reasonable, algebraic steps is an extraneous one. It's a solution to a different equation, that gets, that has a
common intermediate step, but it's also fun to think about why, why this right over here is impossible.