Precalculus (6th Edition) Blitzer

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

Chapter 7 - Section 7.6 - Linear Programming - Exercise Set - Page 874: 38

Answer

$\{-4,-3\}$

Work Step by Step

Step 1. Rearrange the equation as $\sqrt {3x+13}=x+5$ Step 2. Take the square on both sides to get $3x+13=x^2+10x+25$ or $x^2+7x+12=0$ Step 3. Factor the equation to get $(x+3)(x+4)=0$ which gives $x=-4, -3$ Step 4. Check $x=-4$; we have $LHS=\sqrt {3(-4)+13}=1=RHS$ Step 5. Check $x=-3$; we have $LHS=\sqrt {3(-3)+13}=2=RHS$ Step 6. We conclude that the solutions are $x=-4$ and $x=-3$ or $\{-4,-3\}$
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