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

- anonymous

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- phi

set a square root equal to a positive number.

- anonymous

so like 2√(5)

- anonymous

@phi

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## More answers

- 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

- anonymous

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

- anonymous

so 2√(5+4)+2=4

- anonymous

so 2√(5+4)+2=4

- anonymous

@phi

- anonymous

2√(5+4)+2=4
-2 -2
2 √(5+4)=4

- anonymous

2* sqrt 9=6

- anonymous

x=6?

- phi

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

- anonymous

oh right

- anonymous

thank you

- anonymous

ill solve it

- phi

now subtract 2 from both sides
what do you get ?

- anonymous

2√(x+4)=2

- anonymous

do i have to multiply both sides by 2?

- anonymous

@phi

- 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)

- anonymous

ah

- phi

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

- anonymous

√4=1

- anonymous

√x=4=1

- anonymous

√x+4=1

- phi

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

- phi

now square both sides

- anonymous

x+16=1?

- 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"

- anonymous

x+4=1?

- phi

yes

- anonymous

-4 to both side

- phi

last step is add -4 to both sides

- anonymous

x=-3

- anonymous

add 4?

- 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

- anonymous

2√(-3+4)+2=2
2√(-3+4)=0

- anonymous

idk the next step

- anonymous

oh nvm

- anonymous

2√1=0

- 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)

- anonymous

2√(-3+4)+2
2√(1)+2

- phi

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

- anonymous

2

- phi

sqrt(1) means what times itself = 1

- anonymous

oh

- phi

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

- phi

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

- anonymous

i see

- phi

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

- anonymous

2*1=1

- anonymous

2

- anonymous

I'm sorry 2*1 is 2

- anonymous

+2=4

- anonymous

nooooo

- anonymous

i mean yes

- anonymous

its not extraneous

- anonymous

thank you so much sir

- 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)

- anonymous

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

- anonymous

ive lerned more now than the past 2 months in algebra

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