Quadratic Equation B6

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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 using the factorization method.

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