Given the equation;
[tex](3x-1)^2-16=0[/tex]We shall start by expanding the parenthesis as follows;
[tex]\begin{gathered} (3x-1)^2=(3x-1)(3x-1) \\ (3x-1)^2=9x^2-3x-3x+1 \\ (3x-1)^2=9x^2-6x+1 \end{gathered}[/tex]The equation can now be re-written as follows;
[tex]\begin{gathered} 9x^2-6x+1-16=0 \\ 9x^2-6x-15=0 \\ \text{Divide all through by a common factor which is 3;} \\ \frac{9x^2}{3}-\frac{6x}{3}-\frac{15}{3}=0 \\ 3x^2-2x-5=0 \\ \text{ Since the coefficient of x}^2\text{ is greater than 1,} \\ \text{ Multiply the constant (-5) by the coefficient of x squared (3)} \\ \text{The constant now takes the value -15} \\ \text{Two factors of -15 when added together to give -2 are,} \\ 3\text{ and -5} \end{gathered}[/tex]Therefore, we now have;
[tex]\begin{gathered} 3x^2-2x-5=0 \\ 3x^2+3x-5x-5=0 \\ 3x(x+1)-5(x+1)=0 \\ (3x-5)(x+1)=0 \\ \text{Therefore;} \\ (3x-5)=0,(x+1)=0 \\ 3x=5,x=-1 \\ x=\frac{5}{3},x=-1 \end{gathered}[/tex]ANSWER:
[tex]\begin{gathered} \text{Smaller value; x}=-1 \\ \text{Larger value; x}=\frac{5}{3} \end{gathered}[/tex]
