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: 101

Answer

$x=8$

Work Step by Step

$\sqrt{3x+1}=2+\sqrt{x+1}$ Square both sides of the equation: $(\sqrt{3x+1})^{2}=(2+\sqrt{x+1})^{2}$ $3x+1=4+4\sqrt{x+1}+x+1$ $3x+1=x+5+4\sqrt{x+1}$ Reorganize the equation: $3x+1-x-5=4\sqrt{x+1}$ $2x-4=4\sqrt{x+1}$ Once again, square both sides of the equation: $(2x-4)^{2}=(4\sqrt{x+1})^{2}$ $4x^{2}-16x+16=16(x+1)$ $4x^{2}-16x+16=16x+16$ Take all terms to the left side: $4x^{2}-16x+16-16x-16=0$ $4x^{2}-32x=0$ Solve this equation by factoring. Take out common factor $4x$: $4x(x-8)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $4x=0$ $x=\dfrac{0}{4}$ $x=0$ $x-8=0$ $x=8$ Check the solutions found by plugging them into the original equation: $x=0$ $\sqrt{3(0)+1}=2+\sqrt{0+1}$ $\sqrt{1}=2+\sqrt{1}$ $1\ne3$ False $x=8$ $\sqrt{3(8)+1}=2+\sqrt{8+1}$ $\sqrt{25}=2+\sqrt{9}$ $5=2+3$ $5=5$ True The final answer is $x=8$
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