College Algebra 7th Edition

Published by Brooks Cole

Chapter 1, Equations and Graphs - Section 1.4 - Solving Quadratic Equations - 1.4 Exercises: 26

Answer

$x=-4\pm\frac{3\sqrt{6}}{2}$

Work Step by Step

We complete the square: $2x^{2}+16x+5=0$ $x^{2}+8x+\frac{5}{2}=0$ $x^{2}+8x=-\frac{5}{2}$ $x^{2}+8x+16=-\frac{5}{2}+16=\frac{27}{2}$ $(x+4)^{2}=\frac{27}{2}$ $x+4=\displaystyle \pm\sqrt{\frac{27}{2}}=\pm\sqrt{\frac{9*3*2}{2*2}}=\pm\frac{3\sqrt{6}}{2}$ $x=-4\pm\frac{3\sqrt{6}}{2}$

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