Question

help:
completing the square:
2/3x^2-2x+7/3=0

Answer 1

multiply each side by 3
\[\large{2x^2-6x+7}=0\]now divide by two\[\large{x^2-3x+\frac{7}{2}}=0\]\[\large{(x-\frac{3}{2})^2-(\frac{3}{2})^2}+\frac{7}{2}=0\]\[\large{(x-\frac{3}{2})^2+\frac{5}{4}=0}\]

Answer 2

is that already finish?

Answer 3

do you need to find the value of x now?

Answer 4

completing the square has been done by lalaly

Answer 5

yes jimmyrep

Answer 6

ok - i'm not very good at equation thing so

(x - 3/2)^2 = -5/4
take square root of both sides:
x - 3/2 = +- sqrt (-5/4)
x = +- sqrt (-5/4) + 3/2

roots are 3/2 + (sqrt5) i/2 or 3/2 - (sqrt5) i/2

Answer 7

sorry - i was trying to use the equation function but messed up !
are you ok with the answer i gave?

Answer 8

sorry, but i cant totally understand it

Answer 9

did you understand it as far as the fourth line?

Answer 10

no

Answer 11

\[\large{(x-\frac{3}{2})^2}=\frac{-5}{4}\]

Answer 12

\[\large{x-\frac{3}{2}}=\sqrt{\frac{-5}{4}}\]

Answer 13

\[x=\frac{3}{2} \pm \sqrt{\frac{-5}{4}}\]

Answer 14

are you ok with lalaly's last 3 posts?

Answer 15

\[\underline{\large{Hope so}}\]