Chapter 5, Problem 12GR

Solution

To calculate : The solution of the given quadratic equation by factoring.

The solution of the given quadratic equation are $x=1\text{ and }2$ .

Given information :

The quadratic equation is: $3x^2-9x+6=0$ .

Calculation :

Consider the given quadratic equation $3x^2-9x+6=0$ .

Take $3$ as common, so the equation will be

  $\begin{array}{c}3\left(x^2-3x+2\right)=0\\ x^2-3x+2=0\end{array}$

Take the Left Hand Side of the equation, that is:

  $x^2-3x+2$

Find the product of $1\text{ and }2$ .

  $1\times 2=2$

Find the factors of $2$ .

Factors of $2$	Sum of factors
$1,2$	$1+2=3$
$-1,-2$	$-1+\left(-2\right)=-3$

So, rewrite $-3x$ as $-x-2x$ and solve as below:

  $\begin{array}{c}{x}^2-3x+2=x^2-x-2x+2\\ =x\left(x-1\right)-2\left(x-1\right)\\ =\left(x-2\right)\left(x-1\right)\end{array}$

Now, substitute the factors of $x^2-3x+2$ in the given equation as below:

  $\left(x-2\right)\left(x-1\right)=0$

Use zero product property,

  $x-2=0$ or $x-1=0$

Solve for $x$ .

  $\begin{array}{c}x-2=0\\ x=2\end{array}$

Or

  $\begin{array}{c}x-1=0\\ x=1\end{array}$

Now, substitute $x=2$ in the given quadratic equation $3x^2-9x+6=0$ .

  $\begin{array}{c}3{\left(2\right)}^2-9\left(2\right)+6=0\\ 12-18+6=0\\ 0=0\end{array}$

Now, substitute $x=3$ in the given quadratic equation $x^2+6x-27=0$ .

  $\begin{array}{c}3{\left(1\right)}^2-9\left(1\right)+6=0\\ 3-9+6=0\\ 0=0\end{array}$

Therefore, the required solutions are $x=1\text{ and }2$ .