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Mathematics 20 Online
OpenStudy (anonymous):

Part 1. Create two radical equations: one that has an extraneous solution, and one that does not have an extraneous solution. Use the equation below as a model. ax+b+c=d Use a constant in place of each variable a, b, c, and d. You can use positive and negative constants in your equation. Part 2. Show your work in solving the equation. Include the work to check your solution and show that your solution is extraneous. Part 3. Explain why the first equation has an extraneous solution and the second does not.

OpenStudy (anonymous):

@SolomonZelman

OpenStudy (anonymous):

@undeadknight26

OpenStudy (anonymous):

need help plz

undeadknight26 (undeadknight26):

Sorry mate i cannot help you at the moment! @agent0smith @Wolfboy @rational @elementwielder @satellite73 @FlvsGirl

OpenStudy (anonymous):

k when can u

undeadknight26 (undeadknight26):

Thats a hard question XD

OpenStudy (anonymous):

k

OpenStudy (anonymous):

k

OpenStudy (anonymous):

do u know anyone who can

OpenStudy (agent0smith):

Post a screenshot, because any equation of the form ax+b+c=d where a, b, c, and d are all real number constants, always has only one real solution.

OpenStudy (anonymous):

screen shot of what

OpenStudy (agent0smith):

What do you think...? The question.

OpenStudy (anonymous):

i copy and pasted it

OpenStudy (anonymous):

Part 1. Create two radical equations: one that has an extraneous solution, and one that does not have an extraneous solution. Use the equation below as a model. ax+b+c=d Use a constant in place of each variable a, b, c, and d. You can use positive and negative constants in your equation. Part 2. Show your work in solving the equation. Include the work to check your solution and show that your solution is extraneous. Part 3. Explain why the first equation has an extraneous solution and the second does not.

OpenStudy (anonymous):

@agent0smith

OpenStudy (agent0smith):

Ugh.

OpenStudy (anonymous):

can u plz help

OpenStudy (agent0smith):

Copy and paste is NOT a screenshot.

OpenStudy (anonymous):

what the difference

OpenStudy (anonymous):

on computer cant screen shot

OpenStudy (agent0smith):

The difference is pointed out in my first post, go back and read it.

OpenStudy (anonymous):

cant screenshot on computer

OpenStudy (agent0smith):

That doesn't make sense.

OpenStudy (anonymous):

how u screenshot when u on a computer

OpenStudy (agent0smith):

Windows Start menu => type "snipping tool" into the search bar. Or google how to take a screenshot

OpenStudy (anonymous):

OpenStudy (agent0smith):

See how much of a difference that makes? huge.

OpenStudy (anonymous):

k can u help me know

OpenStudy (agent0smith):

Start by putting in some numbers.

OpenStudy (anonymous):

i dont know what to do at all

OpenStudy (agent0smith):

Pick some numbers for a, b, c, d.

OpenStudy (anonymous):

2579

OpenStudy (anonymous):

now what

OpenStudy (anonymous):

can u plz help me plz do it and label ill be back on in a couple hours plz i really need this

OpenStudy (anonymous):

@agent0smith

OpenStudy (anonymous):

\[ax+b+c=d\] has no radical in it my guess is it is \[\sqrt{ax+b}+c=d\]

OpenStudy (anonymous):

you could try \[\sqrt{x+8}+2=5\] which has one nice solution \[\sqrt{x+8}+2=5\\ \sqrt{x+8}=3\\ x+8=9\\ x=1\]

OpenStudy (anonymous):

for one that has no solution you could try the same thing only make it \[\sqrt{x+8}+2=-1\]

OpenStudy (anonymous):

what is for what part

OpenStudy (anonymous):

the second one has no solution try it you will also get \(x=1\) but \(1\) is not a solution to the original equation

OpenStudy (anonymous):

look at this its the picture of what i need

OpenStudy (anonymous):

@satellite73

OpenStudy (anonymous):

@camerondoherty

OpenStudy (camerondoherty):

im nowt good at math sry

OpenStudy (anonymous):

do u no who can help me

OpenStudy (camerondoherty):

5.08?

OpenStudy (anonymous):

yah

OpenStudy (camerondoherty):

oh xD havent finished tht yet lol

OpenStudy (anonymous):

i have 1 and 3 only need 2

OpenStudy (camerondoherty):

i can help u with the first question...

OpenStudy (anonymous):

k

OpenStudy (camerondoherty):

Inser numbers for the a,b,c,d,e ax√x+c+d=e ax√x+c+d=e Like this 2x√x+1+4=2 (For Extaneous Solution) 2x√x+1+2=4 (For Non-Extraneous Solution)

OpenStudy (camerondoherty):

???

OpenStudy (camerondoherty):

Any Questions?

OpenStudy (anonymous):

there no e

OpenStudy (camerondoherty):

?

OpenStudy (anonymous):

u put x+c+d=e but no e its suppose to equal d

OpenStudy (camerondoherty):

well excuse me for trying to help...

OpenStudy (anonymous):

i not being mean just letting u know

OpenStudy (camerondoherty):

all u gotta do is take out tht +1 on each my mistake sry

OpenStudy (anonymous):

why is one Extraneous Solution and other is not

OpenStudy (anonymous):

@AngelWilliams16 @eddielikestacos @gabs15 @Thatoneguyyoulove @deshawn1 @girlnotonfire @victorthagod @adrynicoleb @kathrynturner can u guys help me plz

OpenStudy (anonymous):

deshawn do u know how to do this i need help badly

OpenStudy (anonymous):

@quirah @QuestionsWolf @QUISE94 @Data_LG2 @HisOnlyForever @Natalia8 @kohai @missy22 @Book_Nerd_101 @backwoods15

OpenStudy (anonymous):

@Phebe

OpenStudy (phebe):

im here

OpenStudy (anonymous):

can u help me open this this is what i need help with

OpenStudy (phebe):

sorry it kinda hard @wio can u help

OpenStudy (anonymous):

k thx

OpenStudy (phebe):

wlc

OpenStudy (anonymous):

@flabberghastor

OpenStudy (anonymous):

@helpval22 @hugsnkisses

OpenStudy (anonymous):

@math-geek

OpenStudy (anonymous):

yes?

OpenStudy (anonymous):

can u help me with

OpenStudy (anonymous):

do you have two radical equations?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

they give you a model of \[a \sqrt{x+b}+c=d\] which is a to the square root of x+b then +c which should =d

OpenStudy (anonymous):

they also say to use a constant of a,b,c,and d they should be positive and negative numbers in your equation

OpenStudy (anonymous):

ok i have no idea on how to do it i need alot of help

OpenStudy (anonymous):

give me one sec while i figure it out, then ill explain

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

ok... a radical equation has a square root in the equation...its considered to have an extraneous solution if the final answer does not actually work in the original equation

OpenStudy (anonymous):

that tis not the answer though

OpenStudy (anonymous):

can u make it for me its worth 100 points and i dont get it and if u do ill fan u and give u help with whatever u need i just dont get this lesson plz do it and label part 1 2 3 plz and thx

OpenStudy (anonymous):

r u doing alg online

OpenStudy (anonymous):

yes... but i have till the 13 to finish the corse

OpenStudy (anonymous):

i have till 16

OpenStudy (anonymous):

what r u up to

OpenStudy (anonymous):

what lesson

OpenStudy (anonymous):

plz help me and ill help u

OpenStudy (anonymous):

@wio

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