Question

Solve by completing the square.

2a^2-2a-1=0

Answer 1

first of all, 2a^2 can't even be squared rooted, so how is solving this possible?

Answer 2

Firstly make the coefficient of a^2 = 1..
So divide both the sides by 2 first..

Answer 3

O.O

Answer 4

"2a^2 can't even be squared rooted"

sqrt2 a

Answer 5

okay the whole point of this is to completely square it right? and thats possible if it's x^2+4x+4 because x^2 can be rooted, and 4 can be rooted.

but what i'm supposed to be doing here is ax^2+bx+c=0 then you do this: ax^2+bx+(b/2)^2=c+(b/2)^2

Answer 6

\[( \sqrt{2}x)^{2}-2.\sqrt{2}\sqrt{1/2}+\sqrt{1/2}^{2}-1/2+1=0\]

Answer 7

therefore it should be \[(\sqrt{2}x-\sqrt{1/2})^{2}+1/2\]

Answer 8

uhm i'm not sure if you know what i'm talking about... or you're doing a completely different formula

Answer 9

(b/2)^2 you know that?!

Answer 10

@Steeben didnt get your question??

Answer 11

Square (sqrt2 a -1) and compare it to what you have....

Answer 12

sorry it should be a instead of x. My mistake!!

Answer 13

\[a^2 - a = \frac{1}{2}\]

\[a^2 - a + \frac{1}{4} = \frac{1}{2} + \frac{1}{4}\]

\[(a - \frac{1}{2})^2 = \frac{3}{4} \implies (a - \frac{1}{2})^2 = (\frac{\sqrt{3}}{2})^2\]

Answer 14

I have like... no idea what you guys are talking about. the formula i'm supposed to follow is to do this.


ax^2+bx+c=0

then you add whatever b is to (b/2)^2
and send c to the other side

so it's like this: ax^2+bx+(b/2)^2=-c+(b/2)^2
then you solve, and it'll be a complete square. then you solve for two solutions.

Answer 15

ie Square (sqrt2 a -1)

Answer 16

I have just done the same there..

Answer 17

waterineyes equation is better. You should follow that one.

Answer 18

okayokay i see you divided 2 over haah.
but how did you get the 1/4?

Answer 19

See first tell me what is the coefficient of a after dividing entirely by 2??

Answer 20

\[a ^{2}-2a1/2+(1/2)^{2}-(1/2)^{2}-1=0\]

Answer 21

We get 1/4 while completing the square

Answer 22

well, a^2-a-1/2=0

Answer 23

See, in case of Completion of square method we should two things:

1. Coefficient of \(a^2\) must be 1 if it is not then make it 1..

2. Check for the coefficient of a there, firstly half the coefficient of a and then square it,
Here a has coefficient \(-1\) so divide by 2 to half it that is \(\frac{-1}{2}\)...

3. Now, square of what you get that is \((\frac{1}{2})^2\) so it becomes 1/4 now add this term to both sides of the equation...

Answer 24

ohhh i seeeeee LOL

Answer 25

Got or not??

Answer 26

okay I got some so far, how did you get the \[\sqrt{3}\]

Answer 27

-___- not that....

Answer 28

i mean (rad3)/2)^2

Answer 29

how did you get this: