## anonymous 3 years ago Can someone solve this quadratic equation x^2/3-x-5/3?

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1. anonymous

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

2. anonymous

holy crap use perenthesis

3. anonymous

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

4. anonymous

if so, start by multiplying everything by $$3$$ to get $x^2-3x-5=0$

5. anonymous

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

6. anonymous

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

7. anonymous

so you have to use the quadratic formula

8. anonymous

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

9. anonymous

@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?

10. anonymous

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

11. anonymous

12. anonymous

Okay thats fine

13. anonymous

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 $x=\frac{3\pm\sqrt{27}}{2}$

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. anonymous

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