Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 57: 122

Answer

$x=\pm\sqrt{7}$

Work Step by Step

$\sqrt{11-x^{2}}-\dfrac{2}{\sqrt{11-x^{2}}}=1$ Multiply the whole equation by $\sqrt{11-x^{2}}$: $\sqrt{11-x^{2}}\Big(\sqrt{11-x^{2}}-\dfrac{2}{\sqrt{11-x^{2}}}=1\Big)$ $(\sqrt{11-x^{2}})^{2}-2=\sqrt{11-x^{2}}$ $11-x^{2}-2=\sqrt{11-x^{2}}$ Simplify the left side: $9-x^{2}=\sqrt{11-x^{2}}$ Square both sides: $(9-x^{2})^{2}=(\sqrt{11-x^{2}})^{2}$ $81-18x^{2}+x^{4}=11-x^{2}$ Take all terms to the left side: $x^{4}-18x^{2}+x^{2}+81-11=0$ $x^{4}-17x^{2}+70=0$ Solve this equation by factoring: $(x^{2}-7)(x^{2}-10)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x^{2}-7=0$ $x^{2}=7$ $x=\pm\sqrt{7}$ $x^{2}-10=0$ $x^{2}=10$ $x=\pm\sqrt{10}$ Only $x=\sqrt{7}$ and $x=-\sqrt{7}$ make the initial equation true. The final answer is: $x=\pm\sqrt{7}$
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