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

Jun 21, 2015

Use the quadratic formula to find zeros $x = \frac{3}{2}$ or $x = \frac{6}{5}$

$10 {x}^{2} - 27 x + 18 = \left(2 x - 3\right) \left(5 x - 6\right)$

#### Explanation:

$f \left(x\right) = 10 {x}^{2} - 27 x + 18$ is of the form $a {x}^{2} + b x + c$, with $a = 10$, $b = - 27$ and $x = 18$.

The discriminant $\Delta$ is given by the formula:

$\Delta = {b}^{2} - 4 a c = {27}^{2} - \left(4 \times 10 \times 18\right) = 729 - 720$

$= 9 = {3}^{2}$

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

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{27 \pm 3}{20}$

That is:

$x = \frac{30}{20} = \frac{3}{2}$ and $x = \frac{24}{20} = \frac{6}{5}$

Hence $f \left(x\right) = \left(2 x - 3\right) \left(5 x - 6\right)$

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

Jun 21, 2015

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

#### Explanation:

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

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

Factorising by splitting the middle term

$10 {x}^{2} - 15 x - 12 x + 18 = 0$
$5 x \left(2 x - 3\right) - 6 \left(2 x - 3\right) = 0$
$\textcolor{red}{\left(5 x - 6\right) \left(2 x - 3\right)} = 0$

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

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