Question

i need help writing an equation with one extraneous solution.can someone help me?

Answer 1

Whats your equation

Answer 2

i just need help writing an equaionwith one extraneous solution.i stink at writting those

Answer 3

ok but like what kind of equasion. poloynomial, lond division. or just any equasion

Answer 4

any equation.

Answer 5

45 - 4 + 2y + 4y + 4(y + 7) = -4 - 12y - 4 + 3(y + 2) + 2(2y - 5)
41 + 6y + 4y + 28 = -8 - 12y + 3y + 6 + 4y - 10
69 + 10y = -12 - 5y
81 + 15y = 0
81 = -15y
-81 / 15 = y
-27 / 5 = y
y = -5.4

Answer 6

is this what you are looking for

Answer 7

can you explain to me how to write it?that way if i ne in the future i will know how

Answer 8

and yea that is

Answer 9

well its all there. you write it as in the first line. its just combining like terms and simplyifing

Answer 10

oh ok thank you very much

Answer 11

no problem man. if you want i can write a detailed description out for you and attach it as a file

Answer 12

that would be awsome man.and if i gave you my email could you email it to me?

Answer 13

yeah no problem

Answer 14

ok my email is lawma2898@yahoo.com

Answer 15

cool i send it ovet in a couple mins

Answer 16

thanks man youre a life saver and maybe once i turn 18 i will see you in the marines ;-)

Answer 17

There is no extraneous solution to that equation... it's just a conditional linear equation with one solution..... unless I'm missing something?

Answer 18

what would you suggest man?

Answer 19

There are lots of equation types that give rise to an extraneous solution. A pretty easy one is an equation with a square root expression:

\(\sqrt{2-x}=x\)

Solve by squaring both sides.
\(2-x=x^2\) when you solve this quadratic (do you know how?) you will get 2 solutions. But only ONE of them is a true solution to the original square root equation, because when you check them both back in the original equation (NOT the quadratic, but the original equation), you find that only one of them works.

But most importantly, do you see why the equation given above does NOT work? There is one solution, and it's a "true" solution. Nothing extraneous.

And I'm not a "man". lol

Answer 20

oh...sorry i thought you were cause of the pic.what is your name so i can thank you properly

Answer 21

LOL no problem, I get that a lot. Name is Debbie, the picture is my two sons. :)

Answer 22

could you also explain how to sole it debbie?math is my "trouble" subject

Answer 23

I will, but after I am done, you should really come up with your OWN similar example, so that you know you understand it. :)

ok, after squaring both sides of the original equation we have:

\(2-x=x^2\)
rearrange it:
\(x^2+x-2=0\)
factor it:
\((x+2)(x-1)=0\)
So if a product is=0, then one of the factors is =0, so we have:
\(x+2=0\) so \(x=-2\)
or
\(x-1=0\) so \(x=1\)

Those are both solutions to the quadratic. Now check both solutions back in the original square root equation, \(\sqrt{2-x}=x\):

if \(x=1\):
\(\sqrt{2-1}=\sqrt{1}=1=x\) so that one checks out.

if \(x=-2\):
\(\sqrt{2-(-2)}=\sqrt{4}=2\neq x\) so that one is extraneous.

Answer 24

i will come up with my own dont worry.i dont copy other peoples work

Answer 25

If you are willing to try cooking up your own example, I would start from the factored quadratic and work backwards. E.g., start with something LIKE (but with different solutions than -2 and 1):

\((x+2)(x-1)=0\) multiply it with FOIL:
\(x^2+x-2=0\) rearrange so the squared term is alone:
\(x^2=-x+2\)
\(x^2=2-x\) and take the square root on both sides:
\(x=\sqrt{2-x}\)

That's the process to "cook it up". :)

Answer 26

Good, you'll learn the most that way! :)

Answer 27

ok so when i start writing it if i were to use 5 for the number it would look like\[\sqrt{5+x}\]=x?

Answer 28

Yes, that one will work. Just understand, the quadratic that you end up with in that one is NOT going to factor, which doesn't mean that it doesn't fit the problem but only that SHOWING the solutions and that one is extraneous will be messier.

That's why I suggested that you START from the FACTORED form of the quadratic. :)

Answer 29

how woulod i do that?this whole subject confuses me

Answer 30

I showed you above.

I started with \((x+2)(x-1)=0\)

just start with \((x\pm a)(x\pm b)=0\)

Pick your a, your b, and the signs. then go from there.

Answer 31

oh ok thank you debbie :-) your awsome!!

Answer 32

Make sure at least one of your factors has a "+" or you might end up with 2 real solutions and nothing extraneous. lol