Chapter 7.2, Problem 41WE

To determine

Toshow:The solutions of the given equation are reciprocals.

Answer to Problem 41WE

Thesolution of the given equation is

  $\begin{array}{l}{x}_1=\frac{1}{3}+i\frac{\sqrt{32}}{6}\\ x_2=\frac{1}{3}-i\frac{\sqrt{32}}{6}\end{array}$ .

The solutions of the equation are reciprocals of each other.

Explanation of Solution

Given:

The equation is

  $3x^2-2x+3=0$

Calculation:

For a quadratic equation in the form of $a{x}^2+bx+c=0$ , the solution of this equation is given by

  $x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Comparing the given equation with the standard quadratic equation, the coefficients are:

  $\begin{array}{l}a=3\\ b=-2\\ c=3\end{array}$

Substituting these in the formula for the roots.

First take the positive sign.

  $\begin{array}{c}{x}_1=\frac{-\left(-2\right)+\sqrt{{\left(-2\right)}^2-4\times 3\times \left(3\right)}}{2\times 3}\\ =\frac{2+\sqrt{4-36}}{6}\\ =\frac{1}{3}+i\frac{\sqrt{32}}{6}\end{array}$

Take the negative sign.

  $\begin{array}{c}{x}_2=\frac{-\left(-2\right)-\sqrt{{\left(-2\right)}^2-4\times 3\times \left(3\right)}}{2\times 3}\\ =\frac{2-\sqrt{4-36}}{6}\\ =\frac{1}{3}-i\frac{\sqrt{32}}{6}\end{array}$

Therefore, the solution of the given equation is

  $\begin{array}{l}{x}_1=\frac{1}{3}+i\frac{\sqrt{32}}{6}\\ x_2=\frac{1}{3}-i\frac{\sqrt{32}}{6}\end{array}$ .

Therefore, the solution of the equation are reciprocals of each other.

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