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 X to the
one half and the square root of X mean exactly the same thing.
So we are now ready to use our understanding of roots and exponents to solve algebraic
equations that are a little trickier than the ones we looked at previously.
Let’s start with a 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 X, it’s the quantity X 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 X by itself.
We can’t subtract nine from both sides and expect to get the square root of X, because
the nine is inside the square root, so we would just get the square root of X 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 X plus nine equals twenty-five.
To really prove that squaring a 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 X to the A to the B equals
X to the A times B, and here, A times B, 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.
X plus nine equals twenty-five, so we subtract nine from both sides and we get X equals sixteen.
How about something a little trickier.
The quantity X 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 a 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 X plus two equals four, so we just subtract two from both sides to
get X equals two.
When we have something like X 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 X minus one, and on the right we just lose the radical and get
X plus one.
Then we distribute the nine to get nine X minus nine.
Subtract X, add nine, divide by eight to get ten over eight, simplify by dividing the top
and bottom by two, and X 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 a matter of applying them strategically to undo all the operations that are acting
on the variable you want to solve for.
To get just a little more practice, let’s check comprehension.