Plz help with these questions to see if im right

- geekfromthefutur

Plz help with these questions to see if im right

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- geekfromthefutur

we have to determine if the equation has real or imaginary solutions by solving

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- geekfromthefutur

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- geekfromthefutur

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- geekfromthefutur

@jim_thompson5910 @surjithayer

- jim_thompson5910

show what you have so far

- geekfromthefutur

Well i only have the first one and i said it was a real soultions

- jim_thompson5910

yes that first one definitely has 2 real solutions. Were you able to find the actual solutions?

- TheCatMan

with the 3rd equation start by subtracting 9 to get (x^2=-6)

- geekfromthefutur

yea well i didi have it written down

- jim_thompson5910

ok show us your work please

- TheCatMan

the soluion would come out to a decimal if you work the rest out thats your hint for question 3

- geekfromthefutur

x= ± 2 21−−√ 7
x≈1.30930734,−1.30930734

- geekfromthefutur

thats all i havee for number 1

- geekfromthefutur

for number two i have this hold on

- geekfromthefutur

x=±i6√

- TheCatMan

i say spot one because of the simplification it checks out

- geekfromthefutur

so i think number two is imaginary soutions bevause of the i in the answer

- jim_thompson5910

#2 does have imaginary (aka non real) solutions
but it's not x=±i6√

- geekfromthefutur

What is it

- TheCatMan

then again i am shorting out from the day i had at school with 3 tests so im good for a few helpings tonight

- jim_thompson5910

after you distributed, what did you get?

- geekfromthefutur

i just put it in a math solver

- geekfromthefutur

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- TheCatMan

0 doennt equal -3

- jim_thompson5910

oh, so you're working on #3 then

- geekfromthefutur

nope still 2nd

- TheCatMan

im worked out the second one and got 0=-3 and thats not possible

- jim_thompson5910

i'm referring to this one as the second (since you posted it second)
http://assets.openstudy.com/updates/attachments/56020960e4b0ad22e8a696f9-geekfromthefutur-1442974199899-win_20150922_220854.jpg

- geekfromthefutur

Ohhh sorry my bad yea we are working on the 3rd one my bad

- TheCatMan

square root 6

- geekfromthefutur

How would i write this in two little lines catman

- TheCatMan

sorry my pc just crashed but it restarts fast

- geekfromthefutur

ok

- TheCatMan

you round the decimal to the nearest hundreth

- TheCatMan

so it would be 2.45 instead of 2.449

- geekfromthefutur

ok

- geekfromthefutur

Wait what decimal

- TheCatMan

the square root of 6

- TheCatMan

is not a whole number

- geekfromthefutur

2.44948974278

- TheCatMan

but instead it would be 2.45 as you round the 9 to the 2.44 to a 2.45. note you cannot round up a if the number behind it is less than five for example if the number was 32 it cannot be rounded to 40 it would round down to 30

- geekfromthefutur

ok and what about the next problem

- jim_thompson5910

you mean #3?

- geekfromthefutur

yea

- jim_thompson5910

if you subtract 9 from both sides, you'd get x^2 = -6
taking the square root of both sides leads to what your solver got

- geekfromthefutur

ok x=i,−i

- geekfromthefutur

so imaginary

- jim_thompson5910

no it should be \[\Large x = \pm i*\sqrt{6}\]

- geekfromthefutur

oh ok

- geekfromthefutur

Real quick @jim_thompson5910 this went for this one?

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- jim_thompson5910

oh, I see now
yes, the solution to that one is definitely x = i or x = -i

- geekfromthefutur

So what we went over was for this one

- geekfromthefutur

but the one you said it should be that was for the middle one

- jim_thompson5910

yeah the solution to #2 is x = i or x = -i

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