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vanndaleyrang
Can someone solve this quadratic equation x^2/3-x-5/3?
is it \[\frac{x^2}{3}-x+\frac{5}{3}=0\]?
holy crap use perenthesis
@satellite73 You got it right just change the +5/3 to -5/3
if so, start by multiplying everything by \(3\) to get \[x^2-3x-5=0\]
\[1/3\,{x}^{2}-x-5/3\] that is what maple translates it to
unfortunately i don't think this one factors, as a matter of fact i am sure it does not
so you have to use the quadratic formula
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
@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?
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
satallite already started in quadratic formula instead
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 \[x=\frac{3\pm\sqrt{27}}{2}\]
altho i am pretty sure you will have to learn completing the square sooner or later :)
except i made a mistake, \(5+\frac{9}{4}=\frac{29}{4}\)
Ok, thank you guys! :) @timo86m @satellite73
in any case if the middle term is odd,(-3\) is odd, then it is easier to use the quadratic formula