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Mathematics 16 Online
OpenStudy (anonymous):

please explain?? what are the solutions of 1/2(2x-3)^2+1=11

jimthompson5910 (jim_thompson5910):

1/2(2x-3)^2+1=11 1/2(2x-3)^2+1-1=11-1 1/2(2x-3)^2=10 2*( 1/2(2x-3)^2 )=2*10 (2x-3)^2 = 20 Can you take it from here?

OpenStudy (anonymous):

ill try

jimthompson5910 (jim_thompson5910):

ok tell me what you get

OpenStudy (anonymous):

im stuck

jimthompson5910 (jim_thompson5910):

where

OpenStudy (anonymous):

I squared the equation

jimthompson5910 (jim_thompson5910):

no, you take the square root of both sides

OpenStudy (anonymous):

oh ok

OpenStudy (anonymous):

so far I have 2x-3=2√ 5

jimthompson5910 (jim_thompson5910):

good

jimthompson5910 (jim_thompson5910):

remember it should be +- (plus/minus) 2√ 5

jimthompson5910 (jim_thompson5910):

since there are 2 solutions to any quadratic

OpenStudy (anonymous):

im not sure what to do next?

OpenStudy (anonymous):

do you add the 3 to the other side

jimthompson5910 (jim_thompson5910):

yes

OpenStudy (anonymous):

so is it 2x=3±2√ 5?

jimthompson5910 (jim_thompson5910):

good

jimthompson5910 (jim_thompson5910):

keep going

OpenStudy (anonymous):

so its x=3/2±√5

jimthompson5910 (jim_thompson5910):

you can write it like that (\(\Large x = \frac{3}{2}\pm\sqrt{5}\)) or you can write it like \[\Large x = \frac{3\pm2\sqrt{5}}{2}\]

OpenStudy (anonymous):

ok thanks sooo much

jimthompson5910 (jim_thompson5910):

yw

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