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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

Sal explains what square-root equations are, and shows an example of solving such an equation and checking for extraneous solutions. Created by Sal Khan and CK-12 Foundation.

## Want to join the conversation?

• Hello, I have a question/concern . My question/concern is: At 5;53, one can see that the square root of 2.25 is 1.5. However, the square root of 2.25 can also be -1.5. If we put, " -1.5=4.5-6", x being equal to 2.25 would work. So why doesn't x=2.25 work? • The original equation is (sqrt)x=2x-6. When you see a radical with no + or - sign before it, we assume that we are only taking the principal root (the positive version). When he plugs that 2.25 into the equation, he can't use -1.5 because of that.
• At or so, Sal multiplies 2x and -6 to get -12x which doubles to -24x. Why? I was understanding everything up till that point. • At , why did he take 25^2 and not -25^2 like he did with the -b part of the abc formula? • So why are there extraneous solutions?

I know it's because a square root function (parent graph) has only positive y-values. But why can't we take both roots? Why can't there be 2 y-values for every x-value? A square root function is basically a 90 degree rotation of a quadratic function, so why does it only get half?

I know it's because we want it to be a function. But why does it need to be a function? And who decides what a function is anyway? Who says a function can only have one y-value for every x-value? • Interesting question. The thing is a function itself is defined as having one output for every input. If there are two or more outputs for any input, then you no longer have a function. Also, as to "who decides it gets to be a function", pretty much any equation with both variables and constants with non-ambiguous solutions (i.e. y = sqrt(x)) is a function. So let's see how that relates to your claim that a square root function is basically a 90 degree rotation of a quadratic equation.

Sadly I can't put pictures here but if you go to Desmos and graph both `y=sqrt(x)` and `y=-sqrt(x)`, you'd see you'd basically get the 90-degree rotated quadratic function. However, now we don't even have a function. There are two values per input. Also, this can't be represented in a single equation/function except for `y=(plus or minus)sqrt(x)`. But that's really just the two equations we just graphed, since plus or minus is not an operator but rather the choice between addition or subtraction. The plus or minus operator cannot be used for functions because it just splits the whole equation into two (although it could be used to find solutions of a function like in the quadratic formula).

So in short, the square-root equation is a function by nature because it returns one value per input. Keeping it as a function makes everything clearer. Using the non-function definition of square root really gives you two equations instead of one, so for simplicity, the square root in math returns a positive value by nature. Extraneous solutions from radical equations exist whenever it is assumed that the principal root can return two values in one function.
• How'd you get 625-4 x 4 x 36/8? • What is the definition of "extraneous" and how do I know that an answer is extraneous? • “Extraneous” means “irrelevant or unrelated to the subject being dealt with”. In math, an extraneous solution is a solution that emerges during the process of solving a problem but is not actually a valid solution. You can only find out whether or not a solution is extraneous by plugging the solution back into the original equation. • why does the 24x become 25x after subtracting the x from both sides around ?   