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

Question

4105 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-21T07:23:51

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

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

$f(x) = 10x^2-27x+18$ 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 · 2015-06-21T07:24:35

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

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

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:

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

How do you solve 10x^2 - 27x + 1810x^2 - 27x + 18?