Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.2 - Systems of Linear Equations in Three Variables - Exercise Set - Page 832: 56


The solution of the expression is $\frac{1}{4}\pm i\frac{\sqrt{7}}{4}$.

Work Step by Step

We have to convert the equation into the general form; subtract ${{x}^{2}}+x$ from both sides to get: $\begin{align} & 3{{x}^{2}}+1-{{x}^{2}}-x={{x}^{2}}+x-{{x}^{2}}-x \\ & 2{{x}^{2}}-x+1=0 \end{align}$ Here, $a=2$ , $b=-1$ and $c=1$. By using the quadratic formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, we get the solution, $\begin{align} & x=\frac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 2 \right)\left( 1 \right)}}{2\left( 2 \right)} \\ & =\frac{1\pm \sqrt{1-8}}{4} \\ & =\frac{1\pm \sqrt{-7}}{4} \end{align}$ Since, ${{i}^{2}}=-1$ , Then, $\begin{align} & x=\frac{1\pm \sqrt{7\left( {{i}^{2}} \right)}}{4} \\ & =\frac{1\pm i\sqrt{7}}{4} \\ & =\frac{1}{4}\pm i\frac{\sqrt{7}}{4} \end{align}$ Now, to check whether the solution is correct or not, put $x=\frac{1}{4}\pm i\frac{\sqrt{7}}{4}$ in the equation and get: $\begin{align} & 3{{\left( \frac{1}{4}\pm i\frac{\sqrt{7}}{4} \right)}^{2}}+1={{\left( \frac{1}{4}\pm i\frac{\sqrt{7}}{4} \right)}^{2}}+\left( \frac{1}{4}\pm i\frac{\sqrt{7}}{4} \right) \\ & 3\left( \frac{1\pm 2i\sqrt{7}-7}{16} \right)+\frac{16}{16}=\frac{1\pm 2i\sqrt{7}-7}{16}+\frac{4\left( 1\pm i\sqrt{7} \right)}{16} \\ & \frac{3\pm 6i\sqrt{7}-21+16}{16}=\frac{\pm 2i\sqrt{7}-6+4\pm i4\sqrt{7}}{16} \\ & \frac{-2\pm 6i\sqrt{7}}{16}=\frac{-2\pm 6i\sqrt{7}}{16} \end{align}$ Thus, the solution of the given expression is $\frac{1}{4}\pm i\frac{\sqrt{7}}{4}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.