anonymous
  • anonymous
Help please! (: Solve each equation by undoing each operation: 2(x+4)^2-7=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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jim_thompson5910
  • jim_thompson5910
I'll get you started \[\large 2(x+4)^2-7=0\] \[\large 2(x+4)^2=7\] \[\large (x+4)^2=\frac{7}{2}\]
jim_thompson5910
  • jim_thompson5910
tell me what you get when you finish that up
anonymous
  • anonymous
do u divide 7 and 2?

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jim_thompson5910
  • jim_thompson5910
no leave that as a fraction if you want an exact answer
jim_thompson5910
  • jim_thompson5910
your next step is to undo the square
anonymous
  • anonymous
Im not sure how to do that:/ sorry! would you get 4x^2=4/2 or would you square root both sides..
jim_thompson5910
  • jim_thompson5910
you would apply the square root to both sides
jim_thompson5910
  • jim_thompson5910
since the square root undoes the square
anonymous
  • anonymous
mmkay, so: \[4+ or - \sqrt{7/2}\]
jim_thompson5910
  • jim_thompson5910
close
anonymous
  • anonymous
im confused..
jim_thompson5910
  • jim_thompson5910
well you had x+4 so to undo the +4, you subtract 4 from both sides
jim_thompson5910
  • jim_thompson5910
which means the 4 should be negative
anonymous
  • anonymous
oh, okay. Thanks!
jim_thompson5910
  • jim_thompson5910
and keep in mind that \[\large \sqrt{\frac{7}{2}} = \frac{\sqrt{7}}{\sqrt{2}}\] \[\large \sqrt{\frac{7}{2}} = \frac{\sqrt{7}\sqrt{2}}{\sqrt{2}\sqrt{2}}\] \[\large \sqrt{\frac{7}{2}} = \frac{\sqrt{7*2}}{\sqrt{2*2}}\] \[\large \sqrt{\frac{7}{2}} = \frac{\sqrt{14}}{\sqrt{4}}\] \[\large \sqrt{\frac{7}{2}} = \frac{\sqrt{14}}{2}\]
anonymous
  • anonymous
Oh my goodness, haha(: thanks!!
jim_thompson5910
  • jim_thompson5910
so \[\large x = -4 \pm \sqrt{\frac{7}{2}}\] turns into \[\large x = -4\pm\frac{\sqrt{14}}{2}\]
jim_thompson5910
  • jim_thompson5910
and finally... \[\large x=-4\pm\frac{\sqrt{14}}{2}\] \[\large x=-\frac{8}{2}\pm\frac{\sqrt{14}}{2}\] \[\large x=\frac{-8\pm\sqrt{14}}{2}\]
jim_thompson5910
  • jim_thompson5910
technically once you've isolated x, you're done...but...some books will want it in certain forms and this is usually a form you'll see the answer in
anonymous
  • anonymous
haha(: wow! thanx! GBU!!
jim_thompson5910
  • jim_thompson5910
yw

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