# Quadratic Equation B6

*Ref.*: module B

This exercise requires understanding the fundamental concepts of quadratic equation and their solution methods. It involves using the factorization method.

Solve the quadratic equation:

\[3x^2-9x -72 = -x^2-9x+9\]

As the first step, we need to convert the equation into its standard form. Thus, we get:

\begin{align*} & 3x^2-9x -72+x^2+9x-9= 0 \\[0.6em] & 3x^2+x^2-9x+9x-72-9 = 0 \\[0.6em] & 4x^2-81 =0 \end{align*}

The left member of the equation, \(4x^2-81\), is a notable product, given by the difference of two squares. A difference of two squares \(a^2-b^2\) can be factorised as \((a+b)(a-b)\).

In this case we have \((2x)^2\) and \(9^2\).

The equation becomes: \[(2x+9)(2x-9)=0\]

The solutions are the values of \(x\) for which \(2x+9= 0\) and \(2x-9 = 0\).

\[2x+9 = 0 \to 2x=-9 \to x = -\frac{9}{2} \] \[2x-9 = 0 \to 2x=9 \to x = \frac{9}{2}\]

The solution to the equation is: \[x = -\frac{\displaystyle 9}{\displaystyle 2}, \quad x = \frac{\displaystyle 9}{\displaystyle 2}\]

## Flashcard

Knowing notable products is essential for solving mathematical problems like equations, and memorizing them helps achieve accurate results efficiently and simplifies complex tasks.

\[(a+b)^2 = a^2+2ab+b^2\] \[(a-b)^2 = a^2-2ab+b^2\] \[a^2-b^2=(a+b)(a-b)\] \[(a+b)^3 = a^3+3a^2b+3ab^2+b^3\] \[(a-b)^3 = a^3-3a^2b+3ab^2-b^3\] \[a^3+b^3 = (a+b)(a^2-ab+b^2) \] \[a^3-b^3 = (a-b)(a^2+ab+b^2) \] \[ a^n + b^n = (a + b)(a^{n-1} – a^{n-2}b + a^{n-3}b^2 – \ldots – ab^{n-2} + b^{n-1}) \] \[ a^n + a^n = 2a^n \]