Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 8 - Radical Functions - Chapter Review Exercises - Page 670: 71

Answer

No solution

Work Step by Step

Given \begin{equation} \sqrt{x-4}=5+\sqrt{3 x}. \end{equation} Start by squaring both sides to eliminate the radical sign on the left hand side. Rearrange the equation and then square both sides to eliminate the radical sign on the right hand side and then solve for $x$. \begin{equation} \begin{aligned} \sqrt{x-4}&=5+\sqrt{3 x}\\ \left(\sqrt{x-4}\right)^{2} & =\left(5+\sqrt{3 x}\right)^{2} \\ x-4& =25+10\sqrt{3 x}+3x\\ -(2x+29) & = 10\sqrt{3 x} \\ \left[ -(2x+29) \right]^{2}& =\left(10\sqrt{3 x}\right)^{2}\\ 4x^2+116x+841&= 300x\\ 4x^2-184x+841&= 0. \end{aligned} \end{equation} We solve the quadratic equation using the quadratic formula: \begin{equation} \begin{aligned} a & =4, b=-184, c=841 \\ x & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\ x & =\frac{184 \pm \sqrt{(-184)^2-4 \cdot 4 \cdot(841)}}{2 \cdot 4} \\ & =\frac{184\pm \sqrt{20400}}{8} \\ & =\frac{184\pm \sqrt{400\cdot 51}}{8} \\ & =\frac{184\pm 20\sqrt{51}}{8} \\ & =\frac{46\pm 5 \sqrt{51}}{2}. \end{aligned} \end{equation} This gives: \begin{equation} x=\frac{46+5 \sqrt{51}}{2},\quad x=\frac{46-5 \sqrt{51}}{2}. \end{equation} Check: \begin{equation} \begin{aligned} \sqrt{\frac{46+5 \sqrt{51}}{2}-4}& \stackrel{?}{=} 5+\sqrt{3 \cdot \frac{46+5 \sqrt{51}}{2}}\\ 6.071 & = 16.071 \quad \text {False} \\ \sqrt{\frac{46-5 \sqrt{51}}{2}-4}& \stackrel{?}{=} 5+\sqrt{3 \cdot \frac{46-5 \sqrt{51}}{2}}\\ 1.071 & = 8.929 \quad \text {False}. \end{aligned} \end{equation} No solution.
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