Question

Can someone solve this quadratic equation x^2/3-x-5/3?

Answer 1

is it
\[\frac{x^2}{3}-x+\frac{5}{3}=0\]?

Answer 2

holy crap use perenthesis

Answer 3

@satellite73 You got it right just change the +5/3 to -5/3

Answer 4

if so, start by multiplying everything by \(3\) to get
\[x^2-3x-5=0\]

Answer 5

\[1/3\,{x}^{2}-x-5/3\] that is what maple translates it to

Answer 6

unfortunately i don't think this one factors, as a matter of fact i am sure it does not

Answer 7

so you have to use the quadratic formula

Answer 8

so complete the square :) Also i think it is a good idea to say expression=0 when you say solve. Other wise we imply you mean = to 0

Answer 9

@timo86m Yeah this is my first time using open study so im trying to my best, lol bare with me. Okay can you show me step by step how to complete the square?

Answer 10

use
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
with \(a=1,b=-3,c=-5\) i would not complete the square for this one

Answer 11

satallite already started in quadratic formula instead

Answer 12

Okay thats fine

Answer 13

if you want to complete the square, which the quadratic formula does for you in one step, you have to start with
\[x^2-3x=5\] then take half of 3, which is \(\frac{3}{2}\) and write
\[(x-\frac{3}{2})^2=5+\frac{9}{4}=\frac{27}{4}\] and so
\[x-\frac{3}{2}=\pm\frac{\sqrt{27}}{2}\] making
\[x=\frac{3\pm\sqrt{27}}{2}\]

Answer 14

altho i am pretty sure you will have to learn completing the square sooner or later :)

Answer 15

except i made a mistake, \(5+\frac{9}{4}=\frac{29}{4}\)

Answer 16

Ok, thank you guys! :) @timo86m @satellite73

Answer 17

in any case if the middle term is odd,(-3\) is odd, then it is easier to use the quadratic formula