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Mathematics 18 Online
OpenStudy (anonymous):

What are the exact solutions of x2 + 3x + 1 = 0?

OpenStudy (anonymous):

a. x = the quantity of 3 plus or minus the square root of 13 all over 2 b. x = the quantity of negative 3 plus or minus the square root of 13 all over 2 c. x = the quantity of 3 plus or minus the square root of 5 all over 2 d. x = the quantity of negative 3 plus or minus the square root of 5 all over 2

ganeshie8 (ganeshie8):

use quadratic formula

OpenStudy (anonymous):

Okay

OpenStudy (anonymous):

\[\frac{ -b \pm \sqrt{-b-4(a)(c)} }{ 2a}\] where \[a ^{2}+b+c=0\]

OpenStudy (anonymous):

Which one's a b and c?

OpenStudy (anonymous):

Your equation is \[x ^{2}+3x+1\] So your values would be a = 1 b =3 c=1 Plug in and solve

OpenStudy (anonymous):

Correction to my quadratic formula for to square the b inside the square root

OpenStudy (anonymous):

\[\frac{ -b \pm \sqrt{-b ^{2}-4(a)(c)} }{ 2(a) }\] sorry bout that

OpenStudy (anonymous):

thats okay

OpenStudy (anonymous):

-3+/- sqrt(-3^2-4)/2

OpenStudy (anonymous):

-3+/- sqrt(13)/2

OpenStudy (anonymous):

so that's the answer?

OpenStudy (anonymous):

\[\frac{ -3\pm \sqrt{-3^{2}-4} }{2}\] Solve under the square root

OpenStudy (anonymous):

You haven't simplified

OpenStudy (anonymous):

I did, its -3 +/ sqrt(13)/2

OpenStudy (anonymous):

3 squared is 9 subtract 4 is 5

OpenStudy (anonymous):

ohhhh okay

OpenStudy (anonymous):

Can you guys help me with a couple other similar questions?

OpenStudy (anonymous):

Simply factorize -.- x^2 + 2x + x + 1 = 0 (x+2)(x+1) = 0 x = -1,-2

OpenStudy (anonymous):

Thats wrong answer @ilmc1 @Mike546

OpenStudy (anonymous):

It's the right answer for the correct equation You used 2x not 3x

OpenStudy (anonymous):

What are the exact solutions of x2 + 3x + 1 = 0?

OpenStudy (anonymous):

well its either -3+/- sqrt(13)/2 or -3+/- sqrt(5)/2

OpenStudy (anonymous):

i wrote 3x = 2x + x

OpenStudy (anonymous):

Just look at my answer once more.. :/

OpenStudy (anonymous):

This is a multiple choice problem that did not ask for factoring, read @ilmc1 first response it shows you the options.

OpenStudy (anonymous):

Ah yes I completely skimmed over that

OpenStudy (anonymous):

But what about my answer ?? it cant be wrong if its not given in options

OpenStudy (anonymous):

i understand what you are saying.. but i have checked this answer many times

OpenStudy (anonymous):

Really? Alright lets go over it we'll see if I am wrong (happens) \[(x^2+2x) (x+1)\]

OpenStudy (anonymous):

\[x(x+2) (x+1)\]

OpenStudy (anonymous):

There is your simplified form

OpenStudy (anonymous):

this is equation, isnt it ? \[x^2 + 3x + 1 = 0\]

OpenStudy (anonymous):

OH MY GOD SORRY BRO !! MY BAD pellet @Mike

OpenStudy (anonymous):

:)

OpenStudy (anonymous):

Nevermind please ! :)

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