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: 105


The roots of the equation ${{y}^{2}}-6y-4=0$ by the use of completing the square method are: $y=\left\{ 3+\sqrt{13}\text{ ,}3-\sqrt{13} \right\}$.

Work Step by Step

In completing the square method, convert the equation into the form of ${{\left( a+b \right)}^{2}}$ or ${{\left( a-b \right)}^{2}}$. We have: ${{y}^{2}}-6y-4=0$ ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ But $a=y$, and $\begin{align} & 2ab=6y \\ & 2b=6 \\ & b=3 \end{align}$ So, ${{b}^{2}}={{3}^{2}}=9$ Now, add 9 to both sides of ${{y}^{2}}-6y-4=0$, $\begin{align} & {{y}^{2}}-6y-4+9=9 \\ & {{y}^{2}}-6y+9=9+4 \\ & {{y}^{2}}-6y+9=13 \end{align}$ But, the equation ${{y}^{2}}-6y+9$ is the complete square of ${{\left( y-3 \right)}^{2}}$, ${{\left( y-3 \right)}^{2}}=13$ Taking the square root on both sides, $y-3=\pm \sqrt{13}$ Therefore, the roots are $y=3+\sqrt{13}\text{ or }y=3-\sqrt{13}$.
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