Chapter 7.4, Problem 2WE

(a)

To determine

Solve given equation.

Answer to Problem 2WE

  $x=16\text{ or x}=11$

Explanation of Solution

Given:

The equation is :

  ${\left(x-7\right)}^2-13\left(x-7\right)+36=0$

Calculation:

Substitute

  $(x-7)=z$ in the equation ${\left(x-7\right)}^2-13\left(x-7\right)+36=0$

  $\begin{array}{l}\Rightarrow z^2-13z+36=0\\ \Rightarrow z^2-9z-4z+36=\mathrm{0........}\left[\begin{array}{l}\text{factor the middle term such that the sum of the factors is the middle term and }\\ \text{the product of the factors is the product of first and the last term}\text{.}\end{array}\right]\\ \Rightarrow z(z-9)-4(z-9)=\mathrm{0...}[\text{distributive property]}\\ \Rightarrow \left(z-9\right)\left(z-4\right)=\mathrm{0..........}[\text{distributive property]}\\ \Rightarrow \left(z-9\right)=0\text{ or }\left(z-4\right)=\mathrm{0...}[ab=0\iff a=0\text{ or b=0]}\\ \Rightarrow z=9\text{ or z}=4\end{array}$ Substitute

  $\begin{array}{l}z=(x-7)\\ \Rightarrow x-7=9\text{ or }x-7=4\\ \Rightarrow x=9+7\text{ or x}=4+\mathrm{7........}[add\text{ 7 each side]}\\ \Rightarrow x=16\text{ or x}=11\end{array}$

(b)

To determine

Solve given equation.

Answer to Problem 2WE

  $x=-\frac{8}{3}\text{ or x}=-1$

Explanation of Solution

Given:

The equation is :

  ${\left(1-3x\right)}^2-13\left(1-3x\right)+36=0$

Calculation:

Substitute

  $\left(1-3x\right)=z$ in the equation ${\left(1-3x\right)}^2-13\left(1-3x\right)+36=0$

  $\begin{array}{l}\Rightarrow z^2-13z+36=0\\ \Rightarrow z^2-9z-4z+36=\mathrm{0........}\left[\begin{array}{l}\text{factor the middle term such that the sum of the factors is the middle term and }\\ \text{the product of the factors is the product of first and the last term}\text{.}\end{array}\right]\\ \Rightarrow z(z-9)-4(z-9)=\mathrm{0...}[\text{distributive property]}\\ \Rightarrow \left(z-9\right)\left(z-4\right)=\mathrm{0..........}[\text{distributive property]}\\ \Rightarrow \left(z-9\right)=0\text{ or }\left(z-4\right)=\mathrm{0...}[ab=0\iff a=0\text{ or b=0]}\\ \Rightarrow z=9\text{ or z}=4\end{array}$ Substitute

  $\begin{array}{l}z=(1-3x)\\ \Rightarrow 1-3x=9\text{ or }1-3x=4\\ \Rightarrow -3x=9-1\text{ or }-3x=4-\mathrm{1........}[subtract\text{ 1 from each side]}\\ \Rightarrow -3x=8\text{ or -3x}=3\\ \Rightarrow x=\frac{8}{-3}\text{ or x}=\frac{3}{-3}\mathrm{....................}[\text{divide each side by -3]}\\ \Rightarrow x=-\frac{8}{3}\text{ or x}=-1\end{array}$

(c)

To determine

Solve given equation.

Answer to Problem 2WE

  $x=\pm 3\text{ or }x=\pm 2$

Explanation of Solution

Given:

The equation is :

  $x^4-13x^2+36=0$

Calculation:

Substitute

  $x^2=z$ in the equation $x^4-13x^2+36=0\Rightarrow {\left(x^2\right)}^2-13\left(x^2\right)+36=0$

  $\begin{array}{l}\Rightarrow z^2-13z+36=0\\ \Rightarrow z^2-9z-4z+36=\mathrm{0........}\left[\begin{array}{l}\text{factor the middle term such that the sum of the factors is the middle term and }\\ \text{the product of the factors is the product of first and the last term}\text{.}\end{array}\right]\\ \Rightarrow z(z-9)-4(z-9)=\mathrm{0...}[\text{distributive property]}\\ \Rightarrow \left(z-9\right)\left(z-4\right)=\mathrm{0..........}[\text{distributive property]}\\ \Rightarrow \left(z-9\right)=0\text{ or }\left(z-4\right)=\mathrm{0...}[ab=0\iff a=0\text{ or b=0]}\\ \Rightarrow z=9\text{ or z}=4\end{array}$ Substitute

  $\begin{array}{l}z=x^2\\ \Rightarrow x^2=9\text{ or }{x}^2=4\\ \Rightarrow x=\pm \sqrt{9}\text{ or }x=\pm \sqrt{4}\mathrm{........}[\text{take square root each side]}\\ \Rightarrow x=\pm 3\text{ or }x=\pm 2\end{array}$