Question

How do you solve by completing the square `a^2 -6a-5=0`?

Answer 1

Solution: `a = 3+sqrt 14 and a= 3- sqrt 14`

Explanation:

`a^2-6 a-5=0 or a^2-6 a =5` or

`a^2-6 a +9 =5 +9` or

`(a-3)^2=14 or a-3 = +-sqrt 14` or

`a = 3+- sqrt 14`

Solution: `a = 3+sqrt 14 and a= 3- sqrt 14` [Ans]

Answer 2

`a = 3+sqrt(14), a=3-sqrt(14)`

Explanation:

When we want to make a square out of a quadratic polynomial, we need to look at the b-coefficient of the polynomial (assuming `0 = ax^2+bx+c`).

We know that `(a-b)^2 = a^2 - 2ab -b^2`.
Therefore our "a" is just going to be a, and b is going to be some number for which :

`-2ab = -6a`
`-2b = -6`
`b = 3`

So our square is going to be `(a-3)^2`. However, the square equals to `a^2 - 6a + 9` and we lack the +9 to complete the square.
Luckily, we are free to add a zero to the equation anytime we want. Like this :

`a^2 - 6a + (9 - 9) - 5 = 0`
`a^2 -6a + 9 - 9 -5 = 0`
`(a-3)^2 -14 = 0`
`(a-3)^2 = 14`

Now we can square the equation, watch out for the plus/minus signs :

`|a-3| = sqrt(14)`
`a-3 = +-sqrt(14)`
`a = 3 +-sqrt(14)`

And we are finished.