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## Integrated math 3

### Course: Integrated math 3 > Unit 7

Lesson 2: Square-root equations- Intro to square-root equations & extraneous solutions
- Square-root equations intro
- Intro to solving square-root equations
- Square-root equations intro
- Solving square-root equations
- Solving square-root equations: one solution
- Solving square-root equations: two solutions
- Solving square-root equations: no solution
- Square-root equations

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# Square-root equations intro

Sal gives an example of how an extraneous solution arises when solving 2x-1=√(8-x).

## Want to join the conversation?

- At4:38: Isn't square root of 9 both -3 and +3?(40 votes)
- Yes, sqrt(9) has two roots.

However, unless there is a minus sign in front of the radical, we use the principle root (the positive root. For example:

sqrt(9) = 3

- sqrt(9) = -3 (the minus in front tells you to use the negative root)

Hope this helps.(55 votes)

- Hello, I am curious why Sal assumed 14/8 was a valid solution. Are there not radical equations with no solutions? I was under the impression that one had to test every solution found in the original equation in case there was a chance that all found solutions were extraneous. Any thoughts?(12 votes)
- Yes there are radical equations with no solution and yes you do have to test every solution, but he probably already knew that the solution was valid, which is why he would tell you to check the solution and try it out so that you can practice it.

Hopefully that answers your question.(3 votes)

- At :39 why is there a 4x in the equation?(7 votes)
- You were wondering why (2x-1)^2=4x^2-4x+1, I suppose? Well, (2x-1)^2=(2x-1)(2x-1). This could also be written as 2x(2x-1)-1(2x-1). Then you just use the distributive property for both sets of parentheses:

2x(2x-1)=4x^2-2x

-1(2x-1)=-2x+1

Then combine like terms: 4x^2-4x+1.

All you have to do is make sure you multiply all the terms from one set of parentheses by all the terms from the other set of parentheses and combine all the like terms.

Your question was so long ago that you might have figured it out already, and there are some other good answers to similar questions now.(2 votes)

- at2:48when Sal talks about the quadratic formula and how he gets the equation written in white is very confusing I don't understand how he got that. Can someone explain it to me?(3 votes)
- It's actually very simple, just plug in the values from the quadratic equation into the quadratic formula:

The quadratic formula: x = (-b ± √(b² - 4ac)) / (2a)

The quadratic equation in the video: 4x² - 3x - 7 = 0

You should know that the general form of a quadratic equation is ax² + bx + c = 0, so in this case, a = 4, b = -3, and c = -7. These are the values we will use for the quadratic formula.

Plugging in the values, we get:

x = (-(-3) ± √((-3)² - 4(4 × -7)) / (2 × 4)

x = (3 ± √(9 - 4(-28)) / 8

x = (3 ± √(121)) / 8

x = (3 ± 11) / 8

The first solution:

x = (3 + 11) / 8

x = 14/8

x = 7/4

The second solution:

x = (3 - 11) / 8

x = -8/8

x = -1 (Which is an extraneous solution to the square-root equation in the video)(6 votes)

- Are all negative answers extraneous?(3 votes)
- Not always. When you square both sides of an equation, you get the same equation that you would have if one side had been negative. (e.g. squaring both sides of 3+x=√(-x) and 3+x=-√(-x) give the same equation). So you end up with two solutions, one to each version of the equation. It's perfectly possible that the valid solution is negative, as in the example equation I gave.(3 votes)

- I'm curious, why isn't it a rule that when squaring we default to comparing only to an absolute value? Stepping back for a minute, I'm wondering if doing so requires some sort of larger math principle or this elusive "proof" thing I've not really grasped (i.e. I know what a proof is, but is that's what's necessary and/or does it hurt us once we go further into math).... I can sort of see how that whatever we do here has to hold up even if we layer tons of other math (and other types of math) on top of it, so that, say, this is just some sub-component of a much larger string of equations or just a large equation, that doing so doesn't prevent us from suddenly being unable to solve something that SHOULD be solvable?

Ok, so I hope that makes sense to someone. Whenever I ask questions like this I feel like my math "footing" isn't secure and I'm not truly grasping some fundamental "why/how" and/or that someone/somewhere/somehow is going to connect some of these larger principle dots for me later and that I just need to go through the routine of learning the how first and the why will come later... other times I feel like I MISSED something...

help?(3 votes) - I thought in the quadratic formula its -b{b^2-4ac}/2a but he put a plus 4... this is confusing me..(2 votes)
- It is because he multiplies the expression by -7, which changes the sign to positive(2 votes)

- Can both solutions be extraneous solutions?(2 votes)
- How come he did not just square both sides from the beginning?(2 votes)
- What if there are no extraneous solutions?(2 votes)
- If there are no extraneous solutions, all solutions that you get are valid.(1 vote)

## 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.