Question

Find a solution set to equation
(equation and answer choices in comments)

Answer 1

\[4x ^{2}-2=2x ^{2}+x+16\]
\[a) \frac{ 9 }{ 4 }-\frac{ 1 }{ 4 }\sqrt{65}, \frac{ 9 }{ 4 }+\frac{ 1 }{ 4 }\sqrt{65}\]
\[b) \frac{ 1 }{ 4 }-\frac{ 1 }{ 4 }\sqrt{65}, \frac{ 1 }{ 4 }+\frac{ 1 }{ 4 }\sqrt{65}\]
\[c) -\frac{ 1 }{ 4 }+\frac{ 1 }{ 4 }\sqrt{65}, \frac{ 1 }{ 8 }+\frac{ 1 }{ 8 }\sqrt{65}\]
\[d) -\frac{ 1 }{ 4 }-\frac{ 1 }{ 4 }\sqrt{65}, -\frac{ 1 }{ 4 }+\frac{ 1 }{ 4 }\sqrt{65}\]
\[e) \frac{ 1 }{ 2 }-\frac{ 1 }{ 2 }\sqrt{65}, \frac{ 1 }{ 2 }+\frac{ 1 }{ 2 }\sqrt{65}\]

Answer 2

+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....

Answer 3

oops that 16 is supposed to be a 6. But yes, thanks for the help! @paki

Answer 4

my pleasure...

Answer 5

@paki I got \[\frac{ 1\pm \sqrt{65} }{ 4 }\] does that mean the answer is b?

Answer 6

http://prntscr.com/7uj2zo

Answer 7

@paki thanks so much!

Answer 8

my pleasure... @clara1223

Answer 9

HHAAAAYYYY