# Chapter 7 - Section 7.6 - Solving Equations with Radicals - 7.6 Exercises: 29

$k=0$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $\sqrt{k^2+2k+9}=k+3 ,$ square both sides of the equation. Use special products and concepts of solving quadratic equations. Finally, do checking of the solution with the original equation. $\bf{\text{Solution Details:}}$ Squaring both sides of the equation results to \begin{array}{l}\require{cancel} \left( \sqrt{k^2+2k+9} \right)^2=(k+3)^2 \\\\ k^2+2k+9=(k+3)^2 .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} k^2+2k+9=(k)^2+2(k)(3)+(3)^2 \\\\ k^2+2k+9=k^2+6k+9 .\end{array} Using the propertis of equality to isolate the variable results to \begin{array}{l}\require{cancel} (k^2-k^2)+(2k-6k)=9-9 \\\\ -4k=0 \\\\ k=\dfrac{0}{-4} \\\\ k=0 .\end{array} Upon checking, $k=0$ satisfies the original equation.

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