Chapter 9.8, Problem 2P

To determine

To calculate: To find the numbers

Answer to Problem 2P

Two numbers can be $2,\frac{1}{2}$ or $-2,-\frac{1}{2}$

Explanation of Solution

Given information: Product of two numbers is $1$ and difference of their squares is $\frac{15}{4}$

Calculation:

Let the two numbers be $x$ and $y$

Given that the p

roduct of two numbers is $1$ and difference of their squares is $\frac{15}{4}$

Thus, equations are given as follows:

  $\begin{array}{l}xy=1------(1)\\ x^2-y^2=\frac{15}{4}----(2)\end{array}$

From equation (1),

  $y=\frac{1}{x}$

Substituting the value in equation (2)

  $\begin{array}{l}{x}^2-{\left(\frac{1}{x}\right)}^2=\frac{15}{4}\\ x^2-\frac{1}{x^2}=\frac{15}{4}\\ x^4-1=\frac{15}{4}{x}^2\\ 4x^4-4=15x^2\\ 4x^4-15x^2-4=0\\ 4x^4-16x^2+x^2-4=0\\ 4x^2\left(x^2-4\right)+1\left(x^2-4\right)=0\\ \left(x^2-4\right)\left(4x^2+1\right)=0\\ x^2=4,-\frac{1}{4}\end{array}$

When $x^2=4$ ,

  $x=\pm 2$

When $x=2$

  $y=\frac{1}{2}$

When $x=-2$

  $y=-\frac{1}{2}$

Conclusion:

Hence, two numbers can be $2,\frac{1}{2}$ or $-2,-\frac{1}{2}$