# Quadratic Equation B6

Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/exercises/quadratic-equation-b-6/
Fetched from algebrica.org test 4596; source modified 2025-03-06T18:05:50.

Solve the [quadratic equations](../quadratic-equations.md) using the [factorization method](../factoring-quadratic-equations.md).

\\[
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](../notable-products.md), 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](../notable-products.md) 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
\\]