Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.4 Quadratic Equations - 1.4 Exercises - Page 122: 54


$\color{blue}{\left\{2-\sqrt3, 2+\sqrt3\right\}}$.

Work Step by Step

Add $1$ to both sides of the equation to obtain: $x^2-4x+1=0$ RECALL: The quadratic equation $ax^2+bx+c=0$ can be solved using the quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ The given equation has: $a=1, b=-4, c=1$ Substitute these values into the quadratic formula to obtain: $x=\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(1)}}{2(1)} \\x=\dfrac{4\pm\sqrt{16-4}}{2} \\x=\dfrac{4\pm\sqrt{12}}{2} \\x=\dfrac{4\pm\sqrt{4(3)}}{2} \\x=\dfrac{4\pm2\sqrt{3}}{2} \\x=2 \pm \sqrt3$ Thus, the solution set is $\color{blue}{\left\{2-\sqrt3, 2+\sqrt3\right\}}$.
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.