If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Solving square-root equations: one solution

CCSS.Math:

## 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.