clara1223
  • clara1223
Find a solution set to equation (equation and answer choices in comments)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
clara1223
  • clara1223
\[4x ^{2}-2=2x ^{2}+x+16\] \[a) \frac{ 9 }{ 4 }-\frac{ 1 }{ 4 }\sqrt{65}, \frac{ 9 }{ 4 }+\frac{ 1 }{ 4 }\sqrt{65}\] \[b) \frac{ 1 }{ 4 }-\frac{ 1 }{ 4 }\sqrt{65}, \frac{ 1 }{ 4 }+\frac{ 1 }{ 4 }\sqrt{65}\] \[c) -\frac{ 1 }{ 4 }+\frac{ 1 }{ 4 }\sqrt{65}, \frac{ 1 }{ 8 }+\frac{ 1 }{ 8 }\sqrt{65}\] \[d) -\frac{ 1 }{ 4 }-\frac{ 1 }{ 4 }\sqrt{65}, -\frac{ 1 }{ 4 }+\frac{ 1 }{ 4 }\sqrt{65}\] \[e) \frac{ 1 }{ 2 }-\frac{ 1 }{ 2 }\sqrt{65}, \frac{ 1 }{ 2 }+\frac{ 1 }{ 2 }\sqrt{65}\]
paki
  • paki
+4x^2−2=2x^2+x+16 +4x^2-2x^2-x-2-16=0 can you solve it now...? then use quadratic formula or solve it by factorization....
clara1223
  • clara1223
oops that 16 is supposed to be a 6. But yes, thanks for the help! @paki

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

paki
  • paki
my pleasure...
clara1223
  • clara1223
@paki I got \[\frac{ 1\pm \sqrt{65} }{ 4 }\] does that mean the answer is b?
paki
  • paki
http://prntscr.com/7uj2zo
clara1223
  • clara1223
@paki thanks so much!
paki
  • paki
my pleasure... @clara1223
anonymous
  • anonymous
HHAAAAYYYY

Looking for something else?

Not the answer you are looking for? Search for more explanations.