# Chapter 7 - Section 7.6 - Radical Equations - Exercise Set: 9

$x=3$ or $x=0$

#### Work Step by Step

Square both sides of the equation to obtain: $5x+1=(x+1)^2$ Use the rule $(a+b)^2 = a62 + 2ab+b^2$ where $a=x$ and $b=1$ to obtain: $\\5x+1=x^2+2x+1$ Move all terms to the right side of the equation. Note that when a term is moved to the other side of the equation, its sign changes to its opposite. $0=x^2+2x+1-5x-1 \\0=x^2-3x$ Factor the binomial to obtain: $0=x(x-3)$ Equate each factor to zero then solve each equation to obtain: $x = 0 \text{ or } x -3 = 0 \\x = 0 \text{ or } x = 3$

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