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## Algebra 2

### 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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# Solving square-root equations: one solution

CCSS.Math:

Sal solves the equation 3+√(5x+6)=12. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- What does he mean by "Principle square root"? Is it different from a regular square root?(103 votes)
- If you square a positive number, like '2', you get '4' (positive). If you square the same number in negative form, like '-2', you also get a '4' (positive). So If you take the square root of a '4' you always get a '2' back. What you don't know is whether that '2' was originally a '-2' or a '(positive)2'.

The "Principle square root" means you don't care about the sign, and you are only dealing in the positive domain. So Principle square root of '4' is just '2'.

This becomes important when dealing with roots of variables. Because you literally don't know what the original number is and that is what you are solving for.

Take for example the problem in this video. If he had not mentioned Principle Square root, then your X's answer could be either a negative number or a positive of that number. Plugging the negative or the positive numbers back in the original equation, you would get completely different results.

So if he hadn't said "Principle Square Root", then X could have been either -15 or 15. If you plug in 15 back in the original equation it would check out. But -15 would get all sorts of crazy (try it). So in the end you would've known that the correct answer is 15. But he saved us the trouble of checking the original equation twice by saying "Principle Square Root" or, just the positive answer.(182 votes)

- What types of square roots are there??(6 votes)
- I think what Victoria means is that if there is a principle root, what kind of root is not a principle root? So let me explain. Let's assume x is positive. x^2=x*x , of course. But -x*-x=x^2. You can try this with any number. Let's say x^2=y. So the square root of y can be x or -x. The principle, or positive square root of y is x, but -x is the negative square root.(I'm not sure if you're supposed to say it like that, though. It's just how I say it.)

I hope my explanation did more good than harm. :)(15 votes)

- What does he mean about the negative square root and plus or minus square root at0:23-0:27?(6 votes)
**The plus or minus sign ± means that there are two possible square roots for a number, the positive and the negative square root**

For instance:

±√64=±8

=+8 or -8

Because (+8)^2=64 and (-8)^2=64

Extra explanation:

You might ask... Hey, how is that possible?

Just think about this way, a positive multiplied by a positive will result to a positive and similarly a negative multiplied by a negative will result to a positive. In short if they have the same sign, they will result to positive number.(10 votes)

- What if a square root equals a negative number? Is it unsolvable? For example:

√a(x)+b = -c

a, b, c are definite numbers and x is the variable.

Could you still solve the equation and isolate the variable?(5 votes)- What is the steps for this problem 1/2×+1/4×=9(2 votes)

- How did he get 12 or x=15?(4 votes)
- alright i'll solve that equation

3+(5x+6)^(1/2)=12

to solve it we need to get x alone on one side of the equal sign so to do that we'll subtract 3 from both sides of the equation so the 3 and -3 will cancel eachother on the left side of the equation

(3+(5x+6)^(1/2))-3=12-3

(5x+6)^(1/2)=9

then we need to get rid of the square root over the x so to do that we'll raise both sides of the equation to the power of 2 because 2 is the inverse of 1/2 so when we raise (5x+6)^(1/2) the 2 and 1/2 will be multiplied and cancel eachother

((5x+6)^(1/2))^2=9^2

(5x+6)^((1/2)*2)=9^2

5x+6=9^2

then we do like we did in the first step and subtract by 6 on both sides to move the 6 over to the right side and we'll get

5x=(9^2)-6

and finally we divide by 5 on both sides, then the fives in (5x)/5 will cancel eachother and we'll get

x=((9^2)-6)/5 which is 15

hope that helped(6 votes)

- How can you tell if a radical equation, where the unknown is a radical, has no solution just by looking at it (or is this not possible)?

Through experimentation I have found a way to do this but I'm not sure if it works 100% of the time. Is there no solution if: when the unknown is made the subject of the equation (but still under the square root), and the two sides of the equation have opposite signs?

For example: sqrt(x) = -(5/7)

Will all equations like this ^ have no solution?

Hopefully this makes sense.(3 votes)- In general, when we solve radical equations, we often look for real solutions to the equations. So yes, you are correct that a radical equation with the square root of an unknown equal to a negative number will produce no solution. This also applies to radicals with other even indices, like 4th roots, 6th roots, etc. However, radicals with odd indices (like a cube root) can produce solutions when equal to a negative. For example, the cbrt(x) = -3 has the solution x = -27.(4 votes)

- what do i do after this : x squared (x2) =8(2 votes)
- If you mean x^2 = 8, then you have to transpose the square; look

x^2 = 8

x = sqrt(8)

x = [sqrt(4)][sqrt(2)] // Using the product rule [ sqrt(a)*sqrt(b) = sqrt(ab)] look for perfect squares and simplify

x = 2[sqrt(2)] // sqrt(4) = 2 since 2^2 =4

and

x = -2[sqrt(2)] // sqrt(4) = -2 since (-2)^2 is also equal to 4

Therefore, the final and simplified answer is x = 2[sqrt(2)] and x = -2[sqrt(2)]

*sqrt means square root(3 votes)

- When you put the square root symbol, what's the difference between the symbol for the +/- square root and the princible square root?(2 votes)
- The principal square root is always the positive root. For example, the principal root of sqrt(9) = 3.

But, sqrt(9) has 2 roots. If both roots are needed, usually when solving quadratic equations, then the +/- notation is used as a shorthand for saying we need +sqrt(9) = 3 and -sqrt(9) = -3(3 votes)

- In this video when Sal squares both sides, why doesn't he just multiply both sides by 5x+6?(3 votes)
- On the 5x+6 side this works, but on the 12 side, this does not work because 5x+6 may not equal 12(0 votes)

- Why can't you square both sides right off the bat (to get rid of the square root stuff), and then factor the polynomial that it gives you?(1 vote)
- Problem from the video: 3+√(5x+6)=12

If you try to square both sides right way, the radical will not be eliminate. On the left side, you would have: [ 3+√(5x+6) ]^2

To simplify this, you must use FOIL and it creates:

9 + 3√(5x+6) + 3√(5x+6) + (5x+6) = 5x + 15 + 6√(5x+6)

Notice, we still have a square root.

The only way to make sure the square root is eliminated is to remove everything else from that side. So, Sal subtracted 3 prior to squaring the equation.

Hope this helps.(4 votes)

## Video transcript

We're asked to
solve the equation, 3 plus the principal square root
of 5x plus 6 is equal to 12. And so the general strategy
to solve this type of equation is to isolate the radical sign
on one side of the equation and then you can square
it to essentially get the radical sign to go away. But you have to be
very careful there because when you
square radical signs you actually lose the
information that you were taking the principal
square root. Not the negative square root
or not the plus or minus square root. You are only taking the
positive square root. And so when we get
our final answer, we do have to
check and make sure that it gels with taking
the principal square root. So let's try. Let's see what
I'm talking about. So the first thing
I want to do is I want to isolate this on
one side of the equation. And the best way to isolate
that is to get rid of this 3. And the best way
to get rid of the 3 is to subtract 3 from
the left-hand side. And of course, if I do
it on the left-hand side I also have to do it
on the right-hand side. Otherwise, I would
lose the ability to say that they're equal. And so the left-hand
side right over here simplifies to the principal
square root of 5x plus 6. And this is equal to 12 minus 3. This is equal to 9. And now, we can square both
sides of this equation. So we could square the principal
square root of 5x plus 6 and we can square 9. When you do this-- when you
square this, you get 5x plus 6. If you square the square
root of 5x plus 6, you're going to get 5x plus 6. And this is where we actually
lost some information because we would
have also gotten this if we squared the negative
square root of 5x plus 6. And so that's why we have to be
careful with the answers we get and actually make sure it works
when the original equation was the principal square root. So we get 5x plus 6
on the left-hand side. And on the right-hand
side we get 81. And now, this is just a
straight up linear equation. We want to isolate the x terms. Let's subtract 6
from both sides. On the left-hand side, we have
5x and on the right-hand side, we have 75. And then we can divide
both sides by 5. We get x is equal to--
let's see, it's 15, right? 5 times 10 is 50. 5 times 5 is 25 gives you 75. So we get x is
equal to 15, but we need to make sure
that this actually works for our original equation. Maybe this would
have worked if this was the negative square root. So we need to make
sure it actually works for the
positive square root, for the principal square root. So let's apply it to
our original equation. So we get 3 plus the principal
square root of 5 times 15. So 75 plus 6. So I just took 5
times 15 over here. I put our solution in. It should be equal to 12. Or we get 3 plus square
root of 75 plus 6 is 81 needs to be equal to 12. And this is the principal
root of 81 so it's positive 9. So it's 3 plus 9 needs
to be equal to 12, which is absolutely true. So we can feel pretty
good about this answer.