College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.4 - Solving Quadratic Equations - 1.4 Exercises - Page 123: 53

Answer

$\frac{-2h\pm\sqrt{4h^{2}+2A}}{2}$

Work Step by Step

$A=2x^{2}+4xh$ $2x^{2}+4hx-A=0$ We use the quadratic formula: $x=\frac{-(4h)\pm\sqrt{(4h)^{2}-4(2)(-A)}}{2(2)}=\frac{-4h\pm\sqrt{16h^{2}+8A}}{4}=\frac{-4h\pm\sqrt{4(4h^{2}+2A)}}{4}=\frac{-4h\pm 2\sqrt{4h^{2}+2A}}{4}=\frac{-2h\pm\sqrt{4h^{2}+2A}}{2}$
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