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

Radical and rational equations — Basic example

Watch Sal work through a basic Radical and rational equations problem.

Want to join the conversation?

Video transcript

- [Instructor] What is the solution to the equation above? So we just need to solve for k. So one thing that we could do, well, there's a couple of ways that we could do it. One way is we can multiply both sides of this equation times four k minus three. So let's just do that. Four k minus three. It gets the four k minus three out of the denominator because four k minus three divided by four k minus three. As long as we assume four k minus three isn't equal to zero, that's just going to cancel out and be equal to one. And so this equation is going to simplify to four k minus three times two. And we can actually distribute this two. So this becomes two times four k is eight k. And then two times negative three is negative six. So eight k minus six is equal to, well, all we're left with is 11 over one, or we can just write 11. And so adding six to both sides, you could add six to both sides, and so those add up to zero. You're left with eight k is equal to 17. Now we can just divide both sides by eight, and we get k is equal to 17 over eight. K is equal to 17 over eight.