Precalculus (6th Edition) Blitzer

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

Chapter 1 - Section 1.8 - Inverse Functions - Exercise Set - Page 272: 100


The values of x are $1+\sqrt{2}$ and $1-\sqrt{2}$.

Work Step by Step

Considered the equation, ${{x}^{2}}-2x-1=0$ , Now, follow the steps to solve the equation, 1. Divide both sides by $a$ to make the leading coefficient 1. ${{x}^{2}}-2x-1=0$ Here, the leading coefficient is already 1. 2. Isolate the variable terms on one side of the equation and constant terms on the other side. ${{x}^{2}}-2x=1$ 3. Add the square of half the coefficient of x to both sides of the equation. $\begin{align} & {{x}^{2}}-2x+{{\left( \frac{2}{2} \right)}^{2}}=1+{{\left( \frac{2}{2} \right)}^{2}} \\ & {{x}^{2}}-2x+{{\left( 1 \right)}^{1}}=1+{{\left( 1 \right)}^{1}} \\ & {{x}^{2}}-2x+1=1+1 \\ & {{x}^{2}}-2x+1=2 \end{align}$ Factor the resulting perfect square trinomial, $\begin{align} & {{x}^{2}}-2x+1=2 \\ & {{\left( x-1 \right)}^{2}}=2 \\ \end{align}$ 4. Use the square root property and solve for x. Take the square root of both sides, $\begin{align} & \sqrt{{{\left( x-1 \right)}^{2}}}=\sqrt{2} \\ & x-1=\pm \sqrt{2} \\ & x=1\pm \sqrt{2} \end{align}$ Hence, the values of x are $1+\sqrt{2}$ and $1-\sqrt{2}$.
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