# Complete the square: a^2+2a-3=0.I am stumped as I keep getting the wrong answer please assist.

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Completing the square can unfortunately get quite complex so always try and find the simplest way. The most important fact to remember is that it is the middle term (in this case '2a' )that will determine the outcome.

Take your equation and move the constant to the other side. Thus:

`a^2 + 2a - 3 = 0` becomes

`a^2 +2a = 3`

Now remember we are doing squares so take the `a^2` and place only one in the bracket

(a +....)

Now halve the 2 `(2/2) = 1` from the original middle term (leave the a as it is - always)

(a + 1)

Now put the square back:

`(a+1)^2`

Now remember with equations we can do whatever we like to manipulate them as long as we do the same to the other side. So we have halved the 2 to get 1 and then put in a bracket which is squared. So, do the same on the other side with that same term (ie the original middle term - in our case - 2):

So: `(a+1)^2 = 3 + (2/2)^2`

Remember `(2/2)^2 = 1^2 = 1`

`therefore (a+1)^2 = 3 + 1`

`therefore (a+1)^2 = 4`

Now remove the square by performing the opposite function:

`therefore sqrt((a+1)^2) = sqrt (4)`

`therefore a+1 = ` 2 or -2

`therefore a = 2 - 1` **0r**

`a=-2-1`

`therefore ` **a= 1 or a = -3**

`a^2+2a-3=0`

3 does not make this problem a perfect square trinomial so you add 3 on both sides (because 3 is negative)

`a^2+2a=3`

Now use the formula `(b / 2)^2` to find c `2/2 =1` `1^2 = 1`

The next step is to add one to 3 and make 1 the c

`a^2+2a+1=4` now bring down the sqrt. of a and c, and the middle sign with is between a and b

`(a+1)^2 = 4 `

then use the sqareroot of the parenthesis and 4

`a+1=+-2 ` then solve

`a+1= 2 ` `a+1= -2`

` -1 -1` -1 -1

a=1 a= -3

In case you want to solve this equation : a^2 + 2a -3 = 0

then the solution is given below :

a^2 +2a-3 = 0

=>a^2 +3a -a -3 = 0 [ using middle term factorisation]

=> a(a+3) -1(a+3) =0

=> (a+3)(a-1) =0

Now If (a+3) =0 then a= -3

or if (a-1) =0 then a=1

**Hence solution is a=[-3,1] <--- Answer**