Chapter 7.4, Problem 3WE

(a)

To determine

Solve given equation.

Answer to Problem 3WE

  $y=1\text{ or y}=-\frac{1}{6}$

Explanation of Solution

Given:

The equation is :

  $2{\left(\frac{1}{2y}\right)}^2+5\left(\frac{1}{2y}\right)-3=0$

Calculation:

Substitute

  $\frac{1}{2y}=z$ in the equation $2{\left(\frac{1}{2y}\right)}^2+5\left(\frac{1}{2y}\right)-3=0$

  $\begin{array}{l}\Rightarrow 2z^2+5z-3=0\\ \Rightarrow 2z^2+6z-z-3=\mathrm{0........}\left[\begin{array}{l}\text{factor the middle term such that the sum of the factors is the middle term and }\\ \text{the product of the factors is the product of first and the last term}\text{.}\end{array}\right]\\ \Rightarrow 2z(z+3)-1(z+3)=\mathrm{0...}[\text{distributive property]}\\ \Rightarrow \left(2z-1\right)\left(z+3\right)=\mathrm{0..........}[\text{distributive property]}\\ \Rightarrow \left(2z-1\right)=0\text{ or }\left(z+3\right)=\mathrm{0...}[ab=0\iff a=0\text{ or b=0]}\\ \Rightarrow z=\frac{1}{2}\text{ or z}=-3\end{array}$ Substitute

  $\begin{array}{l}z=\frac{1}{2y}\\ \Rightarrow \frac{1}{2y}=\frac{1}{2}\text{ or }\frac{1}{2y}=-3\\ \Rightarrow 2=2y\text{ or 1}=-3(2y)\mathrm{........}[multiply\text{ 4y and 2y respectively each side]}\\ \Rightarrow y=\frac{2}{2}\text{ or y}=\frac{1}{-6}\mathrm{..............}[\text{divide 2 and -6 respectively each side]}\\ \Rightarrow y=1\text{ or y}=-\frac{1}{6}\end{array}$

(b)

To determine

Solve given equation.

Answer to Problem 3WE

  $y=\pm \frac{3\sqrt{2}}{2}\text{ or }y=\pm 1$

Explanation of Solution

Given:

The equation is :

  $2{\left(y^2-4\right)}^2+5\left(y^2-4\right)-3=0$

Calculation:

Substitute

  $\left(y^2-4\right)=z$ in the equation $2{\left(y^2-4\right)}^2+5\left(y^2-4\right)-3=0$

  $\begin{array}{l}\Rightarrow 2z^2+5z-3=0\\ \Rightarrow 2z^2+6z-z-3=\mathrm{0........}\left[\begin{array}{l}\text{factor the middle term such that the sum of the factors is the middle term and }\\ \text{the product of the factors is the product of first and the last term}\text{.}\end{array}\right]\\ \Rightarrow 2z(z+3)-1(z+3)=\mathrm{0...}[\text{distributive property]}\\ \Rightarrow \left(2z-1\right)\left(z+3\right)=\mathrm{0..........}[\text{distributive property]}\\ \Rightarrow \left(2z-1\right)=0\text{ or }\left(z+3\right)=\mathrm{0...}[ab=0\iff a=0\text{ or b=0]}\\ \Rightarrow z=\frac{1}{2}\text{ or z}=-3\end{array}$ Substitute

  $\begin{array}{l}z=\left(y^2-4\right)\\ \Rightarrow \left(y^2-4\right)=\frac{1}{2}\text{ or }\left(y^2-4\right)=-3\\ \Rightarrow y^2=\frac{1}{2}\text{+4 or }{y}^2=-3+\mathrm{4...........}[\text{add 4 each side ]}\\ \Rightarrow y^2=\frac{9}{2}\text{ or }{y}^2=1\\ \Rightarrow y=\pm \sqrt{\frac{9}{2}}\text{ or }y=\pm \sqrt{1}\mathrm{................}[\text{take square root each side]}\\ \Rightarrow y=\pm \frac{3}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}\text{ or }y=\pm \mathrm{1..........}[\text{rationalize the denominator]}\\ \Rightarrow y=\pm \frac{3\sqrt{2}}{2}\text{ or }y=\pm 1\end{array}$

(c)

To determine

Solve given equation.

Answer to Problem 3WE

  $y=2\text{ or }y=-\frac{1}{3}$

Explanation of Solution

Given:

The equation is :

  $2y^{-2}+5y^{-1}-3=0$

Calculation:

Substitute

  $y^{-1}=z$ in the equation $2y^{-2}+5y^{-1}-3=0$

  $\begin{array}{l}\Rightarrow 2z^2+5z-3=0\\ \Rightarrow 2z^2+6z-z-3=\mathrm{0........}\left[\begin{array}{l}\text{factor the middle term such that the sum of the factors is the middle term and }\\ \text{the product of the factors is the product of first and the last term}\text{.}\end{array}\right]\\ \Rightarrow 2z(z+3)-1(z+3)=\mathrm{0...}[\text{distributive property]}\\ \Rightarrow \left(2z-1\right)\left(z+3\right)=\mathrm{0..........}[\text{distributive property]}\\ \Rightarrow \left(2z-1\right)=0\text{ or }\left(z+3\right)=\mathrm{0...}[ab=0\iff a=0\text{ or b=0]}\\ \Rightarrow z=\frac{1}{2}\text{ or z}=-3\end{array}$ Substitute

  $\begin{array}{l}z=y^{-1}\\ \Rightarrow y^{-1}=\frac{1}{2}\text{ or }{y}^{-1}=-3\\ \Rightarrow \frac{1}{y}=\frac{1}{2}\text{ or }\frac{1}{y}=-\mathrm{3............}[\text{multiply by 2y and y respectively each side ]}\\ \Rightarrow 2=y\text{ or }\frac{\text{ 1}}{-3}=\frac{-3y}{-3}\mathrm{..............}[\text{divide by -3 in the second equation each side]}\\ \Rightarrow y=2\text{ or }y=-\frac{1}{3}\end{array}$