Chapter 7.4, Problem 12WE

To determine

Solve given equation.

Answer to Problem 12WE

  $x\text{=}\pm 2\text{ or }x=\pm 2i$

Explanation of Solution

Given:

The equation is :

  ${\left(x^2+3\right)}^2-6\left(x^2+3\right)=7$

Calculation:

Substitute

  $\left(x^2+3\right)=z$ in the equation ${\left(x^2+3\right)}^2-6\left(x^2+3\right)=7$

  $\begin{array}{l}\Rightarrow z^2-6z=7\\ \Rightarrow z^2-6z-7=\mathrm{0.......................}[\text{subtract 7 from each side]}\\ \Rightarrow z^2-7z+z-7=\mathrm{0........}\left[\begin{array}{l}\text{factor the middle term such that the sum of the factors is the middle term and }\\ \text{the product of the factors is the product of first and the last term}\text{.}\end{array}\right]\\ \Rightarrow z(z-7)+1(z-7)=\mathrm{0...}[\text{distributive property]}\\ \Rightarrow \left(z-7\right)\left(z+1\right)=\mathrm{0..........}[\text{distributive property]}\\ \Rightarrow \left(z-7\right)=0\text{ or }\left(z+1\right)=\mathrm{0...}[ab=0\iff a=0\text{ or b=0]}\\ \Rightarrow z=7\text{ or z}=-1\end{array}$ Substitute

  $\begin{array}{l}z=\left(x^2+3\right)\\ \Rightarrow \left(x^2+3\right)=7\text{ or }\left(x^2+3\right)=-1\\ \Rightarrow x^2=7-3\text{ or }{x}^2=-1-\mathrm{3.......}[subtract\text{ 3 from each side]}\\ \Rightarrow x^2\text{=4 or }{x}^2=-4\\ \Rightarrow x\text{=}\pm \sqrt{4}\text{ or }x=\pm \sqrt{-4}\mathrm{..................}[\text{take square root each side]}\\ \Rightarrow x\text{=}\pm 2\text{ or }x=\pm 2i\mathrm{........................}[\sqrt{-1}=i]\end{array}$

Hence , the solution $x\text{=}\pm 2\text{ or }x=\pm 2i$ .