## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.4 - Product-to-Sum and Sum-to-Product Formulas - Exercise Set - Page 691: 68

#### Answer

The required solution is $\left\{ \frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2} \right\}$

#### Work Step by Step

Let us consider the given expression: ${{u}^{2}}-u-1=0$ The expression is in the form of the quadratic equation. Therefore, by using the formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, the expression can be solved. So, in this question, the value of $a$ is $1$, $b$ is $-1$, and $c$ is $-1$. Now, the expression can be evaluated as shown below: \begin{align} & x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\ & =\frac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\left( 1 \right)} \\ & =\frac{1\pm \sqrt{1+4}}{2} \\ & =\frac{1\pm \sqrt{5}}{2} \end{align} Thus, the possible values are $\frac{1-\sqrt{5}}{2}$ and $\frac{1+\sqrt{5}}{2}$.

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