Question

2 STEP PROBLEM:STEP 1 of 2: 9x^2 = 64 Using the standard form ax^2 + bx + c = 0 of the given quadratic equation, factor the left hand side of the equation into two linear factors. STEP 2 of 2: 9x^2 = 64Solve the quadratic equation by factoring. Write your answer in reduced fraction form, if necessary.

Answer 1

Step 1:

The equation given can be written as:

$9x^2-64=0$

Using the general formula:

$x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

we have that:

$\begin{gathered} x=\frac{-0\pm\sqrt[]{0^2-4(9)(-64)}}{2(9)} \\ x=\frac{\pm\sqrt[]{2304}}{18} \\ x=\frac{\pm48}{18} \\ x=\frac{\pm8}{3} \end{gathered}$

This means that we can factor the equation as:

$\begin{gathered} (x-\frac{8}{3})(x+\frac{8}{3})=0 \\ (3x-8)(3x+8)=0 \end{gathered}$

Therefore the factorization is:

$(3x-8)(3x+8)=0$

Another way to do this is by noticing that the equation:

$9x^2-64=0$

is a difference of squares, using the general formula for difference of squares:

$x^2-y^2=(x-y)(x+y)$

Using this we get the same result as before:

$(3x-8)(3x+8)=0$

Step 2:

From the factorization and using the fact that the multiplication between two numbers can only be zero if one of them is zero we have that:

$\begin{gathered} (3x-8)(3x+8)=0 \\ \text{if and only if} \\ 3x-8=0 \\ or \\ 3x+8=0 \end{gathered}$

Solving each linear equation we have that:

$\begin{gathered} 3x-8=0 \\ 3x=8 \\ x=\frac{8}{3} \end{gathered}$

or

$\begin{gathered} 3x+8=0 \\ 3x=-8 \\ x=-\frac{8}{3} \end{gathered}$

Therefore the solutions of the equation are x=8/3 and x=-8/3