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Mathematics 20 Online
OpenStudy (anonymous):

Solve for x, if it is a complex number, give answer in terms of i 3x^2=2x^2-2x-1 @jim_thompson5910

OpenStudy (anonymous):

This is my last one. I am completely lost.

OpenStudy (anonymous):

@Directrix

jimthompson5910 (jim_thompson5910):

3x^2=2x^2-2x-1 3x^2-2x^2+2x+1 = 0 x^2 + 2x + 1 = 0 I'll let you finish

OpenStudy (anonymous):

I honestly dont know what to do. Do i solve for x?

OpenStudy (anonymous):

If so, it is 1

jimthompson5910 (jim_thompson5910):

yes, but factor first

OpenStudy (anonymous):

(X+1)^2=0

jimthompson5910 (jim_thompson5910):

good, so x = ???

OpenStudy (anonymous):

1?

jimthompson5910 (jim_thompson5910):

no

jimthompson5910 (jim_thompson5910):

Let's test it (x+1)^2=0 (1+1)^2=0 .. plug in x = 1 2^2 = 0 4 = 0 ... this is FALSE, so x = 1 is NOT a solution

OpenStudy (anonymous):

-1

jimthompson5910 (jim_thompson5910):

yep

jimthompson5910 (jim_thompson5910):

Check: (x+1)^2=0 (-1+1)^2=0 .. plug in x = -1 (0)^2 = 0 0 = 0 ... this is TRUE, so x = -1 is definitely a solution

OpenStudy (anonymous):

Sory, had tha the first time, accidentl forgot the negative

OpenStudy (anonymous):

Now, is this a complex #?

jimthompson5910 (jim_thompson5910):

-1 is a real number

OpenStudy (anonymous):

So -1 is the. Answer?

jimthompson5910 (jim_thompson5910):

yes

OpenStudy (anonymous):

Tyvm

jimthompson5910 (jim_thompson5910):

yw

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