Help Stephanie write one radical equation where the solution is extraneous and another equation where the solution is non-extraneous. Using complete sentences, explain each step when solving to justify your examples.

Question

anonymous · Answer

@Yuba  @HelpBlahBlahBlah  @amoodarya  @Awesome781  @denizen  @micahpkay

anonymous · Answer

@Preetha @ParthKohli  @waleedimtiaz  @mustafa2014

anonymous · Answer

@gracerigg @sourwing @zobya @zimmah

anonymous · Answer

somebody please help me out

anonymous · Answer

@TuringTest  @TheKylieeM

turingtest · Answer

a non-extraneous what? what math is this?

anonymous · Answer

algebra 2

anonymous · Answer

and a non-extraneous equation

turingtest · Answer

I had to look it up. You are talking about non-extraneous *solutions*, which is one that satisfies the original equation.

anonymous · Answer

https://www.youtube.com/watch?v=711pdW8TbbY

anonymous · Answer

yeaa

turingtest · Answer

your question is very poorly-worded; you want the "radical of a non-extraneous solution"? 
I don't understand what you want, and since you don't even seem to understand the term "extraneous solution", I'm guessing you don't know either.
Do you have the exact wording of the question in English?

anonymous · Answer

Help Stephanie write one radical equation where the solution is extraneous and another equation where the solution is non-extraneous. Using complete sentences, explain each step when solving to justify your examples

turingtest · Answer

in the future, please put the question exactly in your initial post, you could probably have been helped much earlier

turingtest · Answer

So write an equation involving a $\sqrt x$ somewhere, solve it, then check to see if it's a solution to the initial problem by plugging it back into the beginning. if it is, then it's non-extraneous.

anonymous · Answer

so is this right sqrt x = 2

turingtest · Answer

sure, now solve it

anonymous · Answer

ummm x=4 right

turingtest · Answer

right, now check to see that it solves the original equation

anonymous · Answer

yes it does

turingtest · Answer

so then it must be non-extraneous

anonymous · Answer

ok and for an extraneous solution would this work sqrt x=2x-12

turingtest · Answer

I don;t know, why don't you try it and show me?

anonymous · Answer

idk how to reaally do it what i got so far is x=4x^2-48x+24 then you have to subtract the x from both side which gives you 4x^2-49x+24

anonymous · Answer

@zimmah i looked at the youtube video but im stil confused

anonymous · Answer

do you know that quadratic equation

anonymous · Answer

i think thats the next step

turingtest · Answer

$$ax^2+bx+c=0\implies x={-b\pm\sqrt{b^2-4ac}\over 2a}$$

anonymous · Answer

$$\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }$$

anonymous · Answer

so then its -49 sqrt 2017 over 8

anonymous · Answer

@zimmah @TuringTest am i right so far

turingtest · Answer

remember to write the +/-

turingtest · Answer

and I don't feel like breaking out a calculator to solve the problem this way, so you;ll have to trust your own. just put the numbers in the right places.

anonymous · Answer

i'm not sure how you arrived at that answer, but it doesn't seem to be correct.

Note how you can only use the quadratic formula if on the one side of the equals sign you have 0, and on the other side you have ax^2+bx+c