University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
To unlock all 5,300 videos, start your free trial.
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Solving an equation with a radical and in this case, more than one radical so the process for solving an equation with a radical is almost always the same exact thing. Isolate your radical take both sides to the power to get rid of the radical, solve for your variable and check. The problem in this case we have 2 radicals so we somehow need to isolate two things which we can’t really do.
So what we have to do is just isolate one of the radicals first. So somehow we get one of these radicals by itself. In general I chose to isolate the uglier one the one that’s a little bit more involved so in this case is the 3 x minus 2 but it doesn’t really matter you just need to get one of them by itself.
So I’m going to isolate the first one, so we end up with 3x minus 2 is equal to 1 plus root x plus 3. So now we need to get rid of the square root in order to get rid of the square root we have to square. So I need to square both sides squaring the square root just cancels that out so we are left with 3x minus 2 and then on the other side we are going to have to foil. Make sure you don’t, let’s distribute that 2 into both terms we have to go through a process of foiling it out.
So we have b1 we are going to have a x plus 3 a root x plus 3 and a root x plus 3 which will give us 2, 3x plus 3’s and then the x plus 3 the root x plus 3 times the x plus 3 which is just going to give us x plus 3. Pretty ugly mess what we now want to do is isolate the square root we have one square root left we need to isolate to get it by itself.
So I’m going to get everything else over to this side. So we have a x over here, we need to minus and we have a 1 and a 3 so I’m going to minus 4 from this side as well. Leaving us with 3x minus x just 2x -2 minus 4 is -6 is equal to 2 root x plus 3.
We could solve this out these two radii but I’m going to try to isolate the square root as much as possible so I’m going to divide by that 2 as well make sure we get it completely isolated. Divide by 2 leaving us with x minus 3 is equal to square root of x plus 3.
Let’s continue on over here I’m running out of room. So we have x minus 3 is equal to the square root of x plus 3 this is a problem we know how to solve. We have a square root of isolated equals to something else we need to get rid of the square root and we need to do that by squaring both sides.
So we square the left we square the right when we square the left make sure we follow that out. x² minus 6x plus 9 squaring a square root that just cancels out leaving us with x plus 3. We now need to get everything to one side, subtract x, x² minus 7x subtract 3 plus 6 is equal to 0.
Solving this out we need to factor x minus 6, x minus 1 is equal to 0 leaving us with x is equal to 6, 4, 1. Deep breath so we are able to isolate one radical, square both sides isolate the other radicals square both sides and then get 2 answers. The last thing we have to do is to check these out.
So we got the answers as 1 and 6. Let’s roll back over to our original problem and see that they actually work. So x is equal to 1 we plug in 1, we get to 3 minus 2 we get the square root of 1, minus 1 plus 3 that’s 4 square root of 4. Square root of 1 minus the square root of 4 is 1 minus 2 which is -1. So this side is -1 but we said it is equal to 1. So 1 doesn’t actually work. Checking out our other answer, hopefully that works so we always have some.
X is equal to 6 what we end up getting is 3 time 6 is 18 minus 2 is 16 so the square root of 16, which is 4, minus 6 plus 3, 9 square root f 9 is 3, 4 minus 3 is equal to 1 that works out. So what we’ve ended up doing is getting one answer as x is equal to 6.
It’s a long process hopefully you found me through it but basically all we did was isolate one square root we could have done either 1, I chose the ugly one, square both sides foil it out, isolate the other square root, square both sides, solve take your answer leaving us with a single answer x is equal to 6.
Unit
Roots and Radicals