\[p = \sqrt{2 - p}\]

- anonymous

\[p = \sqrt{2 - p}\]

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- chestercat

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- anonymous

could you better yet show me how to do them?

- anonymous

p=sqrt(2-p)
First, square both sides.
p²=2-p
Then, throw everything to one side.
p²+p-2=0
Then, factor it.
(p+2)(p-1)=0
p+2=0 or p-1=0
p=-2 or p=1

- anonymous

this goes against every rule in algebra so far. my teacher gave every one F's because he thinks we cant learn -_-

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## More answers

- amistre64

sqrt(4) does not equal -2

- anonymous

I wasn't done yet.

- amistre64

1 is good

- anonymous

waiit so what is the other one?

- anonymous

2

- anonymous

p=-2 is wrong, since sqrt(2-p) can't become lower than 2.

- amistre64

only 1; when squareing sides, you run the risk of getting a false answer and have to check it back into the original to see if it works :)

- anonymous

so then it is any number greater than 2?

- amistre64

sqrt(2-p) shouldnt go below zero; below 2 is fine :)

- amistre64

sqrt(2--2) = sqrt(4) = 2
-2 <> 2

- anonymous

ok... this is confusing

- amistre64

what is confusing about it?

- anonymous

just the way my teacher told us only to do it a way i dont remember and like coding learn 2 at once you get raped... :/

- amistre64

there are only two numbers whose value equals their sqrts; do you know those numbers?

- anonymous

for this one?

- amistre64

for any number....
does 0 = sqrt(0) ??
does 1 = sqrt(1) ??

- anonymous

no sqrt for 1
0 = \[\sqrt{0}\]

- amistre64

1*1 = 1 ; therefore; sqrt(1) = 1 right?

- anonymous

wouldnt it equal \[\sqrt{1}\]

- amistre64

1 = 1*1 = sqrt(1) = 1/1 = .....

- amistre64

p = sqrt(2-p) what should we do to solve this?

- anonymous

i dont even know. :/

- amistre64

staring at it wont help :)
Try something, even if its wrong....

- anonymous

give me a minute...

- anonymous

\[-2^{?} +p ^{?}\]

- anonymous

ohh the question marks should be sqrd

- amistre64

its a good attempt; but how do we 'do the same thing' to the other side of the equals?

- amistre64

how do we 'undo' a square root?

- anonymous

square it

- amistre64

yes; so lets try that :) to both sides

- amistre64

p^2 = sqrt(2-p)^2
p^2 = 2-p

- anonymous

4 + psqrd = 0?

- amistre64

do you see that steps I did?

- anonymous

yes

- amistre64

when you square the squarroot; sign; it disappears; that is the only thing that happens to that part....

- anonymous

ok

- amistre64

sqrt(15)^2 = 15
sqrt(3)^2 = 3
sqrt(23654)^2 = 23654
see it?

- amistre64

sqrt(2-p)^2 = 2-p right?

- anonymous

yeah

- amistre64

p^2 = 2-p ; what do you wanna do to this to solve for p?

- anonymous

same side?

- amistre64

ok...lets put it all to the same side....
p^2 +p -2 = 2-2-p+p
p^2 +p -2 = 0

- anonymous

ok

- amistre64

how do we 'unFOIL' this ?

- anonymous

huhm flutter if i know.

- amistre64

please dont use that language here

- anonymous

kk

- anonymous

i never understood the foil. my teacher writes it on the bored and says his thought erases then goes to the next prob

- amistre64

FOIL is a term that is used to teach people how to multiply stuff inside of paranthesis

- anonymous

ok

- amistre64

they teach it in hopes that it will mean something later on; but really its a crutch that keeps you from learning whats really happening to a problem

- amistre64

The distributive property is all that we need; do you remember the distributive property?

- anonymous

i ask for help bet he says a smart alic reply to me... he needs a different career

- anonymous

yeah

- amistre64

How do we use the distributive property then?

- anonymous

times outside the ( ) by the inside of the ( )

- amistre64

very good :)
We can do that as many times as it takes.... watch:
Do you agree that 8*3 = 24?

- anonymous

yep

- amistre64

then 3(5+3) should also equal 24 right?
3(5) + 3(3) = 24
15 + 9 = 24
24 = 24

- anonymous

ok

- amistre64

and... (1+2)(5+3) should also equal 24..like this:
(1+2)(5) + (1+2)(3) = 24
5(1) + 5(2) + 3(1) + 3(2) = 24
5 + 10 + 3 + 6 = 24
15 + 9 = 24
24 = 24
that is what they are calling FOIL..... its just using the distributive property twice

- anonymous

ok
so foil = complete distribution?

- amistre64

yes :)

- anonymous

ok.. u got a mic or something for pc? this would be easier...

- amistre64

nope; using college computers..no sound no video no nothing but a keyboard

- anonymous

darn... ok so please guide me through the problem im about to make a new post... itl be easier...

- amistre64

ok...

- anonymous

i posted it...

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