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Mathematics 13 Online
OpenStudy (anonymous):

Solve for the variable exactly : 1. P^2=2(p+8) 2. 1/n + 2/n-1 =3

OpenStudy (triciaal):

1. distribute to get rid of ( ) rearrange to set = 0 factor solve for x let me know if you need more help / show your values of x 2. is it 2/n or 2/(n-1)? leave the variables on one side when working with fractions multiply everything by the LCD to get rid of the fraction solve for n

OpenStudy (anonymous):

@triciaal for the first problem would I have to do the quadractic formula since there is nothing that multiples to -16 but adds to -2

OpenStudy (anonymous):

& for the second one it was 2/(n-1). I apologize for forgetting the parenthesis. Would the steps still be the same for that one ?

OpenStudy (triciaal):

i got 2+/- rt 132

OpenStudy (triciaal):

yes

OpenStudy (anonymous):

Was that your answer for the first problem? because I got \[ \frac{2 \pm \sqrt{68} }{ 2 }\]

OpenStudy (triciaal):

I have 3n^2-6n -1 =0

OpenStudy (triciaal):

#1 a = 1 yeah my mistake the denominator is 2 but how did you get 68

OpenStudy (triciaal):

4rt8

OpenStudy (triciaal):

1 +/- 2rt8

OpenStudy (anonymous):

\[\frac{ 2 \pm \sqrt{-2^{2}-4(1)(-16)} }{ 2 }\]

OpenStudy (triciaal):

sorry my mistake when I did 4ac I use -2

OpenStudy (anonymous):

So would I be correct in my calculation from before ?

OpenStudy (triciaal):

now simplify what you have

OpenStudy (anonymous):

when i simplified I got \[1 \pm \sqrt{17}\]

OpenStudy (triciaal):

no

OpenStudy (triciaal):

wait

OpenStudy (anonymous):

68 simplifies to 17 * 4. 4 is a perfect sqr, so you take out 2 leaving 17 in the Sqrt, the you can cancel out all the 2's leaving you with one in the numerator

OpenStudy (triciaal):

yes

OpenStudy (triciaal):

that's what i was doing. good job

OpenStudy (anonymous):

thank you for your help

OpenStudy (triciaal):

your welcome

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