Question

solve by completing the square
2x^2 - x + 3 = x + 9

Answer 1

2x^2 = 12

Answer 2

\[2x^2-x-x=9-3\]
\[2x^2-2x=6,x^2-x=3\]
\[add~ both~ sides~ (\frac{ -1 }{2 })^2i.e.,\frac{ 1 }{4 }\]
\[x^2-x+\left( \frac{ 1 }{ 4 } \right)=6+\frac{ 1 }{ 4 }\]
\[\left( x-\frac{ 1 }{2 } \right)^2=\frac{ 25 }{4 }\]
\[x-\frac{ 1 }{2 }=\pm \frac{ 5 }{2 }\]
solve it and get the two values of x.

Answer 3

in x^2 - x+ (1/4) = 6 + 1/4 wouldn't the 6 be a 3?

Answer 4

@surjithayer

Answer 5

you are correct .it should be3+1/4=13/4
\[x-\frac{ 1 }{ 2 }=\pm \frac{ \sqrt{13} }{2 }\]
now you can solve.