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## Algebra II (2018 edition)

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

Lesson 1: Solving square-root equations

# Solving square-root equations: one solution

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? •   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.
• What types of square roots are there?? • 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. :)
• What does he mean about the negative square root and plus or minus square root at - ? • 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.
• 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? • 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. • 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.
• How did he get 12 or x=15? • 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
• 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? • 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.   