Question

How do you solve `y^2-5y-3=0` by completing the square?

Answer 1

`y = 5/2+-sqrt(37)/2`

Explanation:

We will also use the difference of squares identity, which can be written:

`a^2-b^2=(a-b)(a+b)`

with `a=(2y-5)` and `b=sqrt(37)`.

Pre-multiply by `4` to reduce the amount of working in fractions:

`0 = 4(y^2-5y-3)`

`=4y^2-20y-12`

`=(2y-5)^2-25-12`

`=(2y-5)^2-(sqrt(37))^2`

`=((2y-5)-sqrt(37))((2y-5)+sqrt(37))`

`=(2y-5-sqrt(37))(2y-5+sqrt(37))`

Hence:

`y = 5/2+-sqrt(37)/2`

Answer 2

`y = 5.541` OR `y = -0.541`

Explanation:

`y^2 - 5y = 3 rArr` move the constant to the right hand side.

Complete the square by adding what is missing from the square of the binomial to BOTH sides. `(b÷2)^2`

`y^2 - 5y + color(red) (5/2)^2 = 3 + color(red)(5/2)^2`

`(y - 5/2)^2 = 3 + (25/4)`

`y - 5/2 = +-sqrt(9.25` ............`rArr 3+6 1/4 = 9 1/4`

`y = sqrt9.25 +2.5` OR `y = -sqrt9.25 +2.5`

`y = 5.541` OR `y = -0.541`