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: 121

Answer

The real solutions to this equation are $x=\dfrac{1}{2}\pm\dfrac{\sqrt{13}}{2}$

Work Step by Step

$x^{2}\sqrt{x+3}=(x+3)^{3/2}$ Square both sides: $(x^{2}\sqrt{x+3})^{2}=[(x+3)^{3/2}]^{2}$ $x^{4}(x+3)=(x+3)^{3}$ Take $(x+3)$ to divide the right side and simplify: $x^{4}=\dfrac{(x+3)^{3}}{x+3}$ $x^{4}=(x+3)^{2}$ Evaluate the power on the right side: $x^{4}=x^{2}+6x+9$ Take all terms to the left: $x^{4}-x^{2}-6x-9=0$ Factor the left side: $(x^{2}+x+3)(x^{2}-x-3)=0$ Set both factor equal to $0$ and solve each individual equation for $x$: $x^{2}+x+3=0$ Solve using the quadratic formula: $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\dfrac{-1\pm\sqrt{1^{2}-4(1)(3)}}{2(1)}=...$ $...=\dfrac{-1\pm\sqrt{1-12}}{2}=\dfrac{-1\pm\sqrt{-11}}{2}$ Since the number inside the square root is a negative number, solving this equation will yield two complex solutions. Since this problem asks for the real solutions of the equation, move on to the next equation $x^{2}-x-3=0$ Solve using the quadratic formula: $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\dfrac{-(-1)\pm\sqrt{(-1)^{2}-4(1)(-3)}}{2(1)}=...$ $...=\dfrac{1\pm\sqrt{1+12}}{2}=\dfrac{1\pm\sqrt{13}}{2}=\dfrac{1}{2}\pm\dfrac{\sqrt{13}}{2}$ The real solutions to this equation are $x=\dfrac{1}{2}\pm\dfrac{\sqrt{13}}{2}$
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