Chapter 8.7, Problem 2MRE

To determine

To solve the equation.

Answer to Problem 2MRE

The values of $u\text{ is }\frac{2\pm \sqrt{2i}}{3}$

Explanation of Solution

Given information:

  $3u^2-4u+2=0$

Calculation:

The equation can be obtained as

Use Sridhar Acharya formula to solve the equation

  $\begin{array}{l}3u^2-4u+2=0\\ u=\frac{-\left(-4\right)\pm \sqrt{{\left(-4\right)}^2-4\left(3\right)\left(2\right)}}{2\left(3\right)}\left[\text{sridhar acharya formula}\frac{-b\pm \sqrt{b^2-4ac}}{2a}\right]\\ u=\frac{4\pm \sqrt{16-24}}{6}\\ u=\frac{4\pm \sqrt{-8}}{6}\\ u=\frac{4\pm 2\sqrt{2i}}{6}\left[\text{since,}\sqrt{-8}=2\sqrt{2}i\right]\\ u=\frac{2\pm \sqrt{2i}}{3}\left[\text{divide numerator and denominator by 2}\right]\end{array}$

Hence,

The values of $u\text{ is }\frac{2\pm \sqrt{2i}}{3}$