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Mathematics 21 Online
OpenStudy (anonymous):

can someone please come up with a radical equation and check for an extraneous solution?(its not supposed to be extraneous)

OpenStudy (phi):

set a square root equal to a positive number.

OpenStudy (anonymous):

so like 2√(5)

OpenStudy (anonymous):

@phi

OpenStudy (phi):

you want an equation. that means left side = right side you also want a variable (call it x) the idea is to write square root ( some expression with x) = some positive number

OpenStudy (anonymous):

oh and it has to model ax+b+c=d.

OpenStudy (anonymous):

so 2√(5+4)+2=4

OpenStudy (anonymous):

so 2√(5+4)+2=4

OpenStudy (anonymous):

@phi

OpenStudy (anonymous):

2√(5+4)+2=4 -2 -2 2 √(5+4)=4

OpenStudy (anonymous):

2* sqrt 9=6

OpenStudy (anonymous):

x=6?

OpenStudy (phi):

change it to this \[ 2\sqrt{(x+4)}+2=4\] because you need an x

OpenStudy (anonymous):

oh right

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

ill solve it

OpenStudy (phi):

now subtract 2 from both sides what do you get ?

OpenStudy (anonymous):

2√(x+4)=2

OpenStudy (anonymous):

do i have to multiply both sides by 2?

OpenStudy (anonymous):

@phi

OpenStudy (phi):

you want to "get rid" of the 2 in front of the square root notice you have 2 times square root to get rid of the 2, do the opposite of multiply (to both sides)

OpenStudy (anonymous):

ah

OpenStudy (phi):

in other words, divide both sides by 2 what do you get ?

OpenStudy (anonymous):

√4=1

OpenStudy (anonymous):

√x=4=1

OpenStudy (anonymous):

√x+4=1

OpenStudy (phi):

almost, you mean \[ \sqrt{x+4 }=1\]

OpenStudy (phi):

now square both sides

OpenStudy (anonymous):

x+16=1?

OpenStudy (phi):

if you have \[ \sqrt{stuff}\] and you square it, you get back stuff you don't change what's inside .... in other words, the only thing that changes is the square root "goes away"

OpenStudy (anonymous):

x+4=1?

OpenStudy (phi):

yes

OpenStudy (anonymous):

-4 to both side

OpenStudy (phi):

last step is add -4 to both sides

OpenStudy (anonymous):

x=-3

OpenStudy (anonymous):

add 4?

OpenStudy (phi):

yes. and if you put x=-3 into the equation \[ 2 \sqrt{x+4}= 2 \] the left side will simplify to 2, show x=-3 is not extraneous

OpenStudy (anonymous):

2√(-3+4)+2=2 2√(-3+4)=0

OpenStudy (anonymous):

idk the next step

OpenStudy (anonymous):

oh nvm

OpenStudy (anonymous):

2√1=0

OpenStudy (phi):

I put in the wrong original equation you started with \[ 2\sqrt{(x+4)}+2=4 \] to show x=-3 "works" , replace x with -3, and simplify only the left side (leave the = 4 alone)

OpenStudy (anonymous):

2√(-3+4)+2 2√(1)+2

OpenStudy (phi):

yes, and order of operations says now do sqrt(1)

OpenStudy (anonymous):

2

OpenStudy (phi):

sqrt(1) means what times itself = 1

OpenStudy (anonymous):

oh

OpenStudy (phi):

sqrt(1)= 1 (notice 1 is special) another example: sqrt(4) = 2 (because 2*2 is 4)

OpenStudy (phi):

sqrt(1) is 1 because 1*1 is 1

OpenStudy (anonymous):

i see

OpenStudy (phi):

2√(1)+2 becomes 2*1 + 2 now order of operations: do the multiply next

OpenStudy (anonymous):

2*1=1

OpenStudy (anonymous):

2

OpenStudy (anonymous):

I'm sorry 2*1 is 2

OpenStudy (anonymous):

+2=4

OpenStudy (anonymous):

nooooo

OpenStudy (anonymous):

i mean yes

OpenStudy (anonymous):

its not extraneous

OpenStudy (anonymous):

thank you so much sir

OpenStudy (phi):

yes, that is good you just showed \[ 2\sqrt{(x+4)}+2=4 \] becomes 4=4 when x= -3 so x=-3 makes the equation true (left side equals right side)

OpenStudy (anonymous):

Thank you, i just fanned u and gave you a medal and a testimonial

OpenStudy (anonymous):

ive lerned more now than the past 2 months in algebra

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