## Intermediate Algebra (12th Edition)

$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $\sqrt{x^2+12x-4}=x-4 ,$ 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{x^2+12x-4} \right)^2=(x-4)^2 \\\\ x^2+12x-4=(x-4)^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} x^2+12x-4=(x)^2-2(x)(4)+(4)^2 \\\\ x^2+12x-4=x^2-8x+16 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} (x^2-x^2)+(12x+8x)=16+4 \\\\ 20x=20 \\\\ x=\dfrac{20}{20} \\\\ x=1 .\end{array} Upon checking, $x=1$ DOES NOT satisfy the original equation. Hence, there is $\text{ no solution .}$