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Mathematics 14 Online
OpenStudy (anonymous):

solve the equation any way. 2x^2=2x+1 :)

OpenStudy (anonymous):

get the equation= to zero soo subtract both sides by 2x^2

OpenStudy (anonymous):

now factor

OpenStudy (anonymous):

(2x+1)(1x+1)

OpenStudy (anonymous):

wait so how do u factor that?

OpenStudy (anonymous):

@ marie the factorization is incorrect x = 1/2(1 plus-minus root 3)

OpenStudy (anonymous):

oh opps lol thanksand the last step is factoring

OpenStudy (anonymous):

like im really bad at factoring....can u show me how?

OpenStudy (anonymous):

I'm gonna use square root factoring...... I can give you anohter example to factor unless you need to factor for this equation

OpenStudy (anonymous):

Actually this equation becomes a little ugly when we try to factor it, so instead we can use the quadratic equation formula.

OpenStudy (anonymous):

a+or- squareroot of b^2-(a*c) all divided by 2a

OpenStudy (anonymous):

well im suppose to factor it...maybe the quadratic equation would work though.....i have to have a smaller and larger value of x

OpenStudy (anonymous):

if you are supposed to have two differnt answers of x then I would go for quadratic formula

OpenStudy (anonymous):

All right the factors are [x - (1+root3)/2][x - (1-root3)/2]

OpenStudy (anonymous):

so how would u find x through that?

OpenStudy (anonymous):

idk about Aron's way but here's how I would do it. a=-27 b=2 c=1 x= -27+or- squareroot4(-27) all divided by 2(-27)

OpenStudy (anonymous):

the two answers come from one solving for just negative the other solving for just positive

OpenStudy (anonymous):

Just set it equal to 0;the factors [x - (1+root3)/2][x - (1-root3)/2]

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