Need help with completing the sqaure

[Guest]

ok, I'm having trouble completing the square when it comes to fractions.

I'll show you what I did, but I got the wrong answer.

QUESTION: 2x-5+6x^2=0

So I rearanged it: 6x^2 + 2x -5=0

I then divided each term by 6, to get rid of the 6 at the front: x^2+1/3x -5/6=0

then: x^2 + 1/3x-5/6=0
x^2 +1/3x +1/36-1/36-5/6=0
(x+1/6)^2-31/36=0
= 1/6+or - sqaure root of 31/36

BUT the answer is suppose to be (-1+ or - sqare root of 31)/6
so I have an extra number..I got a couple of other completing the questions wrong as well, but maybe it's the same mistake I'm doing here..It would be easier if I could draw these on Msn, as all the / and ^ may be confusing to read for you, and Me, when some1 replies.

-Anna

 

[dagr8est]

6x^2+2x-5=0
6x^2+2x=5
6[x^2+(2/6)x]=5
6[x^2+(1/3)x+1/36]-(6/36)=5
6[(x+1/6)^2]-(6/36)-5=0
6[(x+1/6)^2]-(1/6)-(30/6)=0
6[(x+1/6)^2]-(31/6)=0

I learned completing the square in a different way so I'm not sure where you went wrong, but I hope that helps.

If you want to find the zeroes, use the quadratic formula x=[-b+/-sqrt(b^2-4ac)]/2a Using a quadratic calculator is much easier though. :lol: The zeroes are [-1+sqrt31]/6 and [-1-sqrt31]/6.

 

[Matt]

Hi Anna,

Good work so far.

then: x^2 + 1/3x-5/6=0
x^2 +1/3x +1/36-1/36-5/6=0
(x+1/6)^2-31/36=0
= 1/6+or - sqaure root of 31/36

There should be a minus sign in front of the -1/6, so that you answer reads:

. . . .

Do you see why?

BUT the answer is suppose to be (-1+ or - sqare root of 31)/6

That's absolutely correct. If you add the two fractions together above, then you get:

. . . .

which matches the 'correct' answer. I hope this makes sense!