## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 852: 77

#### Answer

The area of the rectangle is $18$ square units.

#### Work Step by Step

The system of nonlinear equations formed is: \left\{ \begin{align} & {{x}^{2}}+{{y}^{2}}=41 \\ & xy=20 \end{align} \right. We use the substitution method to solve the system of equations. Now, consider Equation (II): \begin{align} & xy=20 \\ & x=\frac{20}{y} \end{align} Substitute the value of x in equation (I) to find the value of y: \begin{align} & {{x}^{2}}+{{y}^{2}}=41 \\ & {{\left( \frac{20}{y} \right)}^{2}}+{{y}^{2}}=41 \\ & \frac{400}{{{y}^{2}}}+{{y}^{2}}=41 \\ & {{y}^{4}}-41{{y}^{2}}+400=0 \end{align} Factorize the above equation to solve for the solutions as shown below: \begin{align} & {{y}^{4}}-41{{y}^{2}}+400=0 \\ & {{y}^{4}}-25{{y}^{2}}-16{{y}^{2}}+400=0 \\ & {{y}^{2}}\left( {{y}^{2}}-25 \right)-16\left( {{y}^{2}}-25 \right)=0 \\ & \left( {{y}^{2}}-25 \right)\left( {{y}^{2}}-16 \right)=0 \end{align} By using the algebraic identity $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$ to simplify the equation, we get: \begin{align} & \left( {{y}^{2}}-25 \right)\left( {{y}^{2}}-16 \right)=0 \\ & \left( {{y}^{2}}-{{5}^{2}} \right)\left( {{y}^{2}}-{{4}^{2}} \right)=0 \\ & \left( y-5 \right)\left( y+5 \right)\left( y-4 \right)\left( y+4 \right)=0 \end{align} Then, the solution of the equation is $y=5,y=-5$, $y=4$, and $y=-4$. Put the value of y in the second equation to find the value of x. For $y=5$, \begin{align} & xy=20 \\ & x\left( 5 \right)=20 \\ & x=\frac{20}{5} \\ & x=4 \end{align} For $y=-5$, \begin{align} & xy=20 \\ & x\left( -5 \right)=20 \\ & x=\frac{20}{-5} \\ & x=-4 \end{align} For $y=4$, \begin{align} & xy=20 \\ & x\left( 4 \right)=20 \\ & x=\frac{20}{4} \\ & x=5 \end{align} For $y=-4$, \begin{align} & xy=20 \\ & x\left( -4 \right)=20 \\ & x=\frac{20}{-4} \\ & x=-5 \end{align} The pairs of solutions for the system of equations is as shown below: $\left\{ \left( 4,5 \right),\left( -4,-5 \right),\left( 5,4 \right),\left( -5,-4 \right) \right\}$. Consider the sides of the rectangle passing through points $\left\{ \left( 4,5 \right),\left( -4,-5 \right),\left( 5,4 \right),\left( -5,-4 \right) \right\}$ is, Length: $\sqrt{{{\left( 5-\left( -4 \right) \right)}^{2}}+{{\left( 4-\left( -5 \right) \right)}^{2}}}=9\sqrt{2}$ Width: $\sqrt{{{\left( 5-4 \right)}^{2}}+{{\left( 4-5 \right)}^{2}}}=\sqrt{2}$ Calculate the area of the rectangle my multiplying the length with the breadth as shown below: $9\sqrt{2}\times \sqrt{2}=18$ Hence, the area of a rectangle is 18 square units.

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