## College Algebra (11th Edition)

$x=\{1,9\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $(x-3)^{2/5}=(4x)^{1/5} ,$ raise both sides to the fifth power to get rid of the denominator in the fractional exponent. Then express the resulting equation in the form $ax^2+bx+c=0,$ and use the concepts of factoring quadratic equations to solve the resulting equation. Finally, do checking if the solution satisfies the original equation. $\bf{\text{Solution Details:}}$ Raising both sides to the fifth power, the given equation becomes \begin{array}{l}\require{cancel} \left((x-3)^{2/5}\right)^5=\left( 4x^{1/5} \right)^5 .\end{array} Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x-3)^{\frac{2}{5}\cdot5 }=(4x)^{\frac{1}{5}\cdot5} \\\\ (x-3)^2=4x .\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)^2-2(x)(3)+(3)^2=4x \\\\ x^2-6x+9=4x \\\\ x^2+(-6x-4x)+9=0 \\\\ x^2-10x+9=0 .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $1(9)=9$ and the value of $b$ is $-10 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -1,-9 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} x^2-x-9x+9=0 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (x^2-x)-(9x-9)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(x-1)-9(x-1)=0 .\end{array} Factoring the $GCF= (x-1)$ of the entire expression above results to \begin{array}{l}\require{cancel} (x-1)(x-9)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x-1=0 \\\\\text{OR}\\\\ x-9=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-1=0 \\\\ x=1 \\\\\text{OR}\\\\ x-9=0 \\\\ x=9 .\end{array} Upon checking, $x=\{1,9\}$ satisfy the original equation.