Question

How do you solve the quadratic `6y^2-13y+6=0` using any method?

Answer 1

Using the quadratic formula
`color(white)("XXX")color(green)(y=2/3color(black)(" or ")y=3/2)`

Explanation:

The quadratic formula for the general quadratic (in `y`)
`color(white)("XXX")ay^2+by+c=0`
is
`color(white)("XXX")y=(-b+-sqrt(b^2+4ac))/(2a)`

For the example: `6y^2-13y+6=0`
`color(white)("XXX")a=6`
`color(white)("XXX")b=-13`
`color(white)("XXX")c=6`
giving
`color(white)("XXX")y=(13+-sqrt((-13)^2-4(6)(6)))/(2(6))`

`color(white)("XXX")=(13+-sqrt(169-144))/12`

`color(white)("XXX")=(13+-5)/12`

`color(white)("XXX")=18/12= 3/2color(black)(" or ")=8/12=2/3`

Answer 2

`y = 2/3` or `y = 3/2`

Explanation:

Use an AC method:

Look for a pair of factors of `AC = 6*6=36` with sum `B=13`

The pair `9, 4` works.

Use this pair to split the middle term, then factor by grouping as follows:

`0 = 6y^2-13y+6`

`color(white)(0) = 6y^2-9y-4y+6`

`color(white)(0) = (6y^2-9y)-(4y-6)`

`color(white)(0) = 3y(2y-3)-2(2y-3)`

`color(white)(0) = (3y-2)(2y-3)`

Hence `y = 2/3` or `y = 3/2`