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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?

• At : Isn't square root of 9 both -3 and +3? •  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.
• 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? • At :39 why is there a 4x in the equation? • 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.
• at when 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? • 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)
• Are all negative answers extraneous? • 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.
• I thought in the quadratic formula its -b{b^2-4ac}/2a but he put a plus 4... this is confusing me.. • How come he did not just square both sides from the beginning? • What if there are no extraneous solutions? • 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?  