Chapter 6.8, Problem 5STP

To determine

To calculate: The solution set for the equation: $3{\left(2x+1\right)}^2=27$ .

Answer to Problem 5STP

The solution of the equation is $3{\left(2x+1\right)}^2=27$ is: $x=-2\text{ or }x=1$ option $B$ is correct.

Explanation of Solution

Given information:

The solution set for the equation: $3{\left(2x+1\right)}^2=27$ .

The options are:

  $\begin{array}{l}\text{A }\left[-5,4\right]\text{ C }\left[2,-1\right]\\ \text{B }\left[-2,1\right]\text{ D }\left[-3,3\right]\end{array}$

Formula used:

When two or more functions are in the form of equation, apply the expanded form of the monomials.

Calculation:

So, the given expression as,

  $3{\left(2x+1\right)}^2=27$

Divide each side by $3$ ,

  $\begin{array}{c}3{\left(2x+1\right)}^2=27\\ \frac{3{\left(2x+1\right)}^2}{3}=\frac{27}{3}\\ {\left(2x+1\right)}^2=9\end{array}$

Now apply the identity here:

  ${\left(a+b\right)}^2=a^2+b^2+2ab$

  $\begin{array}{c}{\left(2x+1\right)}^2=9\\ {\left(2x\right)}^2+{\left(1\right)}^2+2\cdot 2x\cdot 1=9\\ 4x^2+4x+1=9\end{array}$

Now subtract $9$ from both sides:

  $\begin{array}{c}4x^2+4x+1=9\\ 4x^2+4x+1-9=9-9\\ 4x^2+4x-8=0\end{array}$

Factor out the greatest common factor by the method of middle term factorization:

  $\begin{array}{c}4x^2+4x-8=0\\ 4\left(x^2+x-2\right)=0\\ 4\left(x^2+2x-x-2\right)=0\end{array}$

Factor out the greatest common factor:

  $\begin{array}{c}4\left(x^2+2x-x-2\right)=0\\ 4\left(x\left(x+2\right)-1\left(x+2\right)\right)=0\\ 4\left(\left(x+2\right)\left(x-1\right)\right)=0\\ \left(x+2\right)\left(x-1\right)=0\end{array}$

Now either:

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

Since these are the solutions of the function:

  $x=-2,x=1$

Hence, the simplified functions of $3{\left(2x+1\right)}^2=27$ is $x=-2,x=1$ , option $B$ is correct.