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# Extraneous solutions of radical equations

Video transcript

- [Voiceover] So let's say
we have the radical equation two-x minus one is equal to the square root of eight minus x. So we already have the
radical isolated on one side of the equation. So, we might say, "Well, let's
just get rid of the radical. "Let's square both
sides of this equation." So we might say that
this is the same thing as two-x minus one squared is equal to the square
root of eight minus x, eight minus x squared,
and then we would get, let's see, two-x minus one squared is four x-squared minus four-x plus one is equal to eight minus x. Now we have to be very,
very, very careful here. We might feel, "Okay we
did legitimate operations. "We did the same thing to both sides. "That these are equivalent equations." But, they aren't quite equivalent. Because when you're squaring something, one way to think about it, is when you're squaring it,
you're losing information. So, for example, this would be true even if the original
equation were two-x... Let me, this in a different color. Even if the original
equation were two-x minus one is equal to the negative of the square root of eight minus x. Because if you squared both sides of this, you would also get, you would also get that right over there, because a negative squared
would be equal to a positive. So, when we're finding a solution to this, we need to test our solution to make sure it's truly the solution to this first yellow equation here, and not the solution to this up here. If it's a solution to
this right-hand side, and not the yellow one, then we would call that
an extraneous solution. So, let's see if we can solve this. So let's write this as kind
of a standard quadratic. Let's subtract eight from both sides. So let's subtract eight from both sides to get rid of this eight over here, and let's add x to both sides. So, plus x, plus x, and we are going to get, we are going to get four-x squared minus three-x, minus seven, minus seven, is equal to, is equal to zero. And let's see, we would want to factor this right over here, and, let's see, maybe I could do this by, if I do it by... Well, I'll just use the
quadratic formula, here. So the solutions are going to be... X is going to be equal
to negative b, so three, plus or minus the square root of b-squared, so negative
three squared is nine, minus four times a, which is four, times c, which is negative seven. So I could just say times... Well, I'll just write a seven here and then that negative is
gonna make this a positive. All of that over two-a. So two times four is eight. So, this is gonna be three plus or minus the square root of... Let's see, four times
four is 16 times seven. 16 times seven is gonna be 70 plus 42. Let me make sure I'm doing this right. So 16 times seven. The two, four. So, it's 112 plus nine. So, 121, that worked out nicely. So, plus or minus the square root of 121, all of that over eight. Well, that is equal to
three plus or minus 11, all of that over eight. So that is equal to, if we add 11, that is 14-eights. Or, if we subtract 11, three minus 11 is negative eight. Negative eight divided
by eight is negative one. So we have to think about... You might say, "Okay,
I found two solutions "to the radical equation." But remember, one of
these might be solutions to this alternate radical equation that got lost when we squared both sides. We have to make sure
that they're legitimate, or maybe one of these is
an extraneous solution. In fact, one is very likely a solution to this radical equation which
wasn't our original goal. So, let's see. Let's try out x equals negative one. If x equals negative one, we would have two times negative one minus one is equal to
the square root of eight minus negative one. So that would be negative
two minus one is equal to the square root of, is equal to the square root of nine. And so we'd have negative three is equal to the square root of nine. The principle root of nine. This is a positive square root. This is not true. So this, right over here, that is an extraneous solution. Extraneous. Extraneous solution. It is a solution to this
one right over here. Because notice, for that one, if you substitute two times negative one minus one is equal to the negative of eight minus negative one. So this is negative three is equal to the negative of three. So, it checks out for this one. So, this one right over here
is the extraneous solution. This one right over here is
gonna be the actual solution for our original equation, and you can test it out on your own. In fact, I encourage you to do so.