Solving Algebraic Equations With Roots and Exponents

Hey it’s Professor Dave, let’s do some more algebra. We just learned about all kinds of roots. Square roots, cube roots, and more. We learned that they can also be represented with fractional exponents, where to the one half and the square root of mean exactly the same thing. So we are now ready to use our understanding of roots and exponents to solve algebraic equations that are little trickier than the ones we looked at previously. Let’s start with simple one. The square root of (X plus nine) equals five. Notice here that the square root symbol extends over this entire term, so it’s not just nine more than the square root of it’s the quantity plus nine, and then the square root of that whole thing. This means we have to be very specific regarding the order in which we do things to try to get by itself. We can’t subtract nine from both sides and expect to get the square root of because the nine is inside the square root, so we would just get the square root of plus nine, minus nine, and that just makes things worse. Instead, we have to get rid of the square root first. Well we know that the opposite of taking the square root of something would be squaring it. So we can square both sides to get rid of this square root. That gives us plus nine equals twenty-five. To really prove that squaring square root gets rid of the root, let’s remember that taking the square root is the same as raising to the one half power, so we can rewrite this equation like this. On the left, we have this term to the one half power to the second power. We can recall from our rules of manipulating exponents that to the to the equals to the times and here, times or one half times two, equals one. So the one half and the two essentially negate one another to just leave the term by itself. Now things look much simpler. plus nine equals twenty-five, so we subtract nine from both sides and we get equals sixteen. How about something little trickier. The quantity plus two to the three halves power equals eight. We still just go from the outside in, undoing every operation that acts on the variable we are solving for, one at time. First we have to get rid of the exponent. How do we get rid of the three halves? The fastest way to do it would be to raise both sides to the two thirds power. If we raise this term that is to the three halves to the two thirds power, we are raising it to the three halves times two thirds power, and if we do this, everything cancels, and we just get one as the exponent. Another way to do it would be to square both sides and then take the cube root, but why do two steps when you can do one? Now on the right, we have eight to the two thirds. This we can just evaluate by taking the cube root, which is two, and then squaring to get four. Now we are left with plus two equals four, so we just subtract two from both sides to get equals two. When we have something like squared plus five equals forty-one, this is also quite easy, we just subtract five, and then take the square root. But remember, when we take the square root of something we get two answers, in this case positive six and negative six, so we have to keep in mind that these types of equations have two solutions. Let’s do one more. Three root (X minus one) equals root (X plus one). Certainly we must get rid of these square roots, so let’s square both sides. Don’t forget that on the left we square the three and the radical term, so we get nine times the quantity minus one, and on the right we just lose the radical and get plus one. Then we distribute the nine to get nine minus nine. Subtract add nine, divide by eight to get ten over eight, simplify by dividing the top and bottom by two, and will equal five fourths. So as you can see, once you know all the rules for how to manipulate roots and exponents, it’s just matter of applying them strategically to undo all the operations that are acting on the variable you want to solve for. To get just little more practice, let’s check comprehension.