Chapter 8.2, Problem 40PPS

To determine

To solve: the given equation for $n$ . Check the result.

Answer to Problem 40PPS

The value of $n$ in the given equation is $0,-1$ .

Explanation of Solution

Given information:

Given equation

  $2n\left(3n+3\right)=0$

Calculation:

Consider the given equation,

  $2n\left(3n+3\right)=0$

If product of two numbers is 0, then the individual is equals to 0.

  $2n=0\Rightarrow n=0$

And

  $3n+3=0$

Now, solve the equation as shown,

  $\begin{array}{c}3n+3-3=0-3\\ 3n=-3\\ \frac{3n}{3}=\frac{-3}{3}\\ n=-1\end{array}$

Hence, $n=-1$ or $n=0$ .

Now, check the result by back substitution.

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

And

  $\begin{array}{c}2n\left(3n+3\right)=0\\ 2\left(-1\right)\left(3\left(-1\right)+3\right)=0\\ -2\times \left(-3+3\right)=0\\ \left(-2\right)\times 0=0\\ 0=0\end{array}$

Which is true.

Hence, the value of $n$ in the given equation is $0,-1$ .