Question

How do you solve `10x^2 - 27x + 18`?

Answer 1

Use the quadratic formula to find zeros `x = 3/2` or `x=6/5`

`10x^2-27x+18 = (2x-3)(5x-6)`

Explanation:

`f(x) = 10x^2-27x+18` is of the form `ax^2+bx+c`, with `a=10`, `b=-27` and `x = 18`.

The discriminant `Delta` is given by the formula:

`Delta = b^2-4ac = 27^2 - (4xx10xx18) = 729 - 720`

`=9 = 3^2`

Being a positive perfect square, `f(x) = 0` has two distinct rational roots, given by the quadratic formula:

`x = (-b+-sqrt(Delta))/(2a) = (27+-3)/20`

That is:

`x = 30/20 = 3/2` and `x = 24/20 = 6/5`

Hence `f(x) = (2x-3)(5x-6)`

graph{10x^2-27x+18 [-0.25, 2.25, -0.28, 0.97]}

Answer 2

`color(red)(x= 6/5 , x =3/2`

Explanation:

`10x^2 - 27x + 18 = 0`

We can first factorise the above expression and thereby find the solution.

Factorising by splitting the middle term

`10x^2 - 15x - 12x + 18 = 0`
`5x(2x- 3) - 6(2x - 3) = 0`
`color(red)( (5x - 6)(2x- 3) ) = 0`

Equating each of the two terms with zero we obtain solutions as follows:

`color(red)(x= 6/5 , x =3/2`