Question

SOLVE the equation. Use factoring techniques (3x - 1)^2 - 16 = 0X= Smaller value X= Larger value

Answer 1

Given the equation;

`(3x-1)^2-16=0`

We shall start by expanding the parenthesis as follows;

`\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}`

The equation can now be re-written as follows;

`\begin{gathered} 9x^2-6x+1-16=0 \\ 9x^2-6x-15=0 \\ \text{Divide all through by a common factor which is 3;} \\ (9x^2)/(3)-(6x)/(3)-(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}`

Therefore, we now have;

`\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=(5)/(3),x=-1 \end{gathered}`

ANSWER:

`\begin{gathered} \text{Smaller value; x}=-1 \\ \text{Larger value; x}=(5)/(3) \end{gathered}`