Question

Mr. Jones is asked to write two types of radical equations for the SAT Prep Course. The first solution to the radical equation must be extraneous. The second solution to the radical equation must be non-extraneous. Write one equation where the solution is extraneous. Then write a second equation where the solution is non-extraneous. Using complete sentences, explain each step when solving to justify your examples.
Can you please help me out with this question?@jim_thompson5910

Answer 1

@jim_thompson5910

Answer 2

are you familiar with the term "extraneous" ?

Answer 3

isn't is when something is excluded from the original equation? @jim_thompson5910

Answer 4

in a way, yes

Answer 5

okay

Answer 6

there's more to it though

Answer 7

let's say we had this equation

\[\Large \sqrt{x} = -1\]

and we state that x is a real number

Answer 8

what is the solution for that equation?

Answer 9

I don't understand extraneous and non extraneous solutions very well, that is the main reason why i need help

Answer 10

what operation undoes square roots?

Answer 11

if you square something it undoes the square root.

Answer 12

correct

Answer 13

square both sides of the equation I wrote. What do you get?

Answer 14

-1

Answer 15

-1 times -1 is not -1

Answer 16

(-1)^2 = (-1)*(-1) = ?

Answer 17

sorry 1

Answer 18

so x = 1 is a possible solution

Answer 19

we need to check it

Answer 20

\[\Large \sqrt{x} = -1\]

\[\Large \sqrt{1} = -1\]

\[\Large 1 = -1\]

which is false

Answer 21

so x = 1 is considered to be an "extraneous solution"

Answer 22

it pops up in the process of solving, but it does NOT satisfy the original equation

Answer 23

here is a page that has another example

http://www.mathwords.com/e/extraneous_solution.htm

Answer 24

but in the question it says that i also need a non-extraneous solution. @jim_thompson5910

Answer 25

did you check out the link I posted?

Answer 26

it provides an example where 2 solutions pop up...but 1 of those solutions is extraneous

Answer 27

which one is the non-extraneous?

Answer 28

the one that makes the original equation true

Answer 29

check both x = 9 and x = 4 with the original equation

Answer 30

oh okay thanks!! is it okay if we go on twiddla later to check my work? @jim_thompson5910

Answer 31

sure we can do that

Answer 32

thank you!! :)

Answer 33

np

Answer 34

let's find out

Answer 35

so x is what value?

Answer 36

that is the only possible solution for x

Answer 37

you need to check it now

Answer 38

so you are correct in stating that x = 81 is non-extraneous

Answer 39

ok sounds good

Answer 40

@jim_thompson5910 can u plz get on twiddla now?

Answer 41

Lincoln and Gemma are looking at the equation the square root of the quantity of 2 times x minus 3 equals square root of x . Lincoln says that the solution is extraneous. Gemma says the solution is non-extraneous. Is Lincoln correct? Is Gemma correct? Are they both correct? Justify your response by solving this equation, explaining each step with complete sentences @jim_thompson5910

Answer 42

what do you have so far?

Answer 43

nothing i need help with it

Answer 44

@jim_thompson5910

Answer 45

The equation is \[\Large \sqrt{2x}-3 = \sqrt{x}\] right?

Answer 46

yes

Answer 47

what happens when you square both sides

Answer 48

idk i need help with it to understand

Answer 49

\[\Large \sqrt{2x}-3 = \sqrt{x}\]

\[\Large (\sqrt{2x}-3)^2 = (\sqrt{x})^2\]

\[\Large (\sqrt{2x})^2-2*\sqrt{2x}*3+(3)^2 = (\sqrt{x})^2\]

\[\Large 2x-6*\sqrt{2x}+9 = x\]

\[\Large -6*\sqrt{2x}+9 = x-2x\]

\[\Large -6*\sqrt{2x}+9 = -x\]

\[\Large -6*\sqrt{2x} = -x-9\]

\[\Large \sqrt{2x} = \frac{-x-9}{-6}\]

\[\Large \sqrt{2x} = \frac{x+9}{6}\]

I'll let you finish up

Answer 50

sorry I made a typo, but I fixed it

Answer 51

ok

Answer 52

I'm really confused :/ @jim_thompson5910 i'm so sorry

Answer 53

where are you stuck?

Answer 54

everything..

Answer 55

you understand how I squared both sides right? I did so to undo the square root

Answer 56

and I also used the rule (a-b)^2 = a^2 - 2ab + b^2 to expand out the left side

Answer 57

okay then thats probably where i got stuck.. cuz idk that formula @jim_thompson5910

Answer 58

that is the perfect square formula

Answer 59

you would foil out (a-b)(a-b) to get a^2 - 2ab + b^2

Answer 60

similar to (a+b)(a+b) = a^2 + 2ab + b^2

Answer 61

okay so how would i answer the question?

Answer 62

you would keep going to solve for x

Answer 63

can u go on twiddla?

Answer 64

\[\Large \sqrt{2x} = \frac{x+9}{6}\]

\[\Large (\sqrt{2x})^2 = (\frac{x+9}{6})^2\]

\[\Large 2x = \frac{(x+9)^2}{36}\]

Answer 65

do you see how to solve that for x?

Answer 66

no

Answer 67

(x+9)^2 turns into what expression

Answer 68

(x+9)(x+9)?

Answer 69

foil that out

Answer 70

x^2+9x+9x+81?

Answer 71

that turns into x^2 + 18x + 81

Answer 72

\[\Large 2x = \frac{(x+9)^2}{36}\]

\[\Large 2x = \frac{x^2+18x+81}{36}\]

\[\Large 2x*36 = x^2+18x+81\]

\[\Large 72x = x^2+18x+81\]

Answer 73

the ultimate goal is to isolate x

Answer 74

u keep adding stuff and that confuses me where did u get 36 from?

Answer 75

6 square is 36, sry

Answer 76

it's okay i'm just a bit slow math isn't my strongest subject

Answer 77

\[\Large (\frac{x+9}{6})^2 = \frac{(x+9)^2}{6^2} \]

\[\Large (\frac{x+9}{6})^2 = \frac{x^2+18x+81}{36} \]

Answer 78

You're doing quite well actually

Answer 79

not really.. hah i have no clue what's going on atm.

Answer 80

well you do understand that squaring undoes square roots

so that is usually the first thing to remember when it comes to problems like this

Answer 81

the other stuff is just moving terms around, simplifying and such

Answer 82

at first, all of this is very confusing, but with a lot of practice it should come easier

Answer 83

:) thanks for the encouragement! really needed that :)

Answer 84

You're welcome. You might find this hard to believe, but I also struggled with math too. I just kept at it and eventually things clicked. So don't give up.

Answer 85

:) thank you so much!!

Answer 86

see attached

Answer 87

try this