# Quadratic Equation B7
Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/exercises/quadratic-equation-b-7/
Fetched from algebrica.org test 4600; source modified 2025-03-06T18:06:28.
Solve the [quadratic equations](../quadratic-equations.md) using the [factorization method](../factoring-quadratic-equations.md).
\\[
(4x + 8)\\left(\\frac{1}{2}x-6\\right) = 0
\\]
* * *
Expanding the equation using the distributive property, we get:
\\begin{align\*} & \\frac{4}{2}x^2-24x + \\frac{8}{2}x-48 = 0 \\[0.6em] &2x^2-24x + 4x - 48 = 0\\[1em] &2x^2-20x-48 = 0 \\end{align\*}
The coefficients \\(a, b\\) and \\(c\\) have 2 as common multiplier. We can simplify the equation which is now brought into the standard form of a quadratic equation:
\\[
x^2-10x-24 = 0
\\]
* * *
The equations is in the standard form \\(ax^2+bx+c=0\\). It is essential to verify the its [discriminant](../quadratic-formula.md) \\(\\Delta = b^2 - 4ac\\) is \\(\\geq0\\) to ensure the equation admits solutions in the field of real numbers. Substituting the coefficients of the equation into \\(\\Delta\\), we get:
\\[
\\Delta = (-10)^2 - 4(1)(-26) = 100 + 96 = 196 \\geq 0
\\]
\\(\\Delta \\gt 0\\) means the equation has real solutions.
* * *
Now, we need to [factorize](../factoring-quadratic-equations.md) the [polynomial](../polynomials.md). We must find two numbers, \\(r\_1, r\_2\\) whose sum \\(S = r\_1 + r\_2\\) equals \\(b = -10\\) and whose product \\(P = r\_1 \\cdot r\_2\\) equals \\(a \\cdot c = 1 \\cdot -24 = -24 \\). We can use this simple table to find the numbers that satisfy our constraints.
\\begin{array}{rrrr} & r\_1 & r\_2 & P & S \\\\ \\hline & 2 & - 12 & -24 & -10 \\\\ & -2 & +12 & -24 & 10\\\\ \\end{array}
The numbers \\(r\_1, r\_2\\) satisfying the constraint are \\(2\\) and \\(-12\\) (row 1). Then we need to rewrite the polynomial as \\(ax^2 + r\_{1}x + r\_{2}x + c\\).
* * *
The equation becomes:
\\[
x^2+2x-12x-24 = 0
\\]
Factoring common terms, we get:
\\begin{align\*} &x(x+2)-12(x+2) = 0 \\[0.6em] &(x+2)(x-12) = 0 \\end{align\*}
The solutions are the values of \\(x\\) for which \\(x-2= 0\\) and \\(x-12 = 0\\).
\\[
x -2 = 0 \\to x=-2
\\]
\\[
x -12 = 0 \\to x=12
\\]
The solution to the equation is:
The solution to the equation is:
\\[
x\_=-2 \\quad \\quad x\_1 =12
\\]
Remember that the [discriminant](../quadratic-formula.md) is crucial in determining the nature and number of solutions of quadratic equations.
- If \\( b^2 - 4ac > 0\\), the quadratic equation has two distinct real solutions.
\\[
S = \\{x\_1, x\_2\\} \\quad x\_1, x\_2 \\in \\mathbb{R} \\quad x\_1 \\neq x\_2
\\]
- If \\( b^2 - 4ac = 0\\), the quadratic equation has two coincident real solutions.
\\[
S = \\{x\\} \\quad x \\in \\mathbb{R} \\quad x = x\_1 = x\_2
\\]
- If \\( b^2 - 4ac = < 0\\), the quadratic equation has no real solutions. Instead, it gives rise to complex solutions characterized by imaginary components.
\\[
\\nexists \\hspace{10px} x \\in \\mathbb{R}
\\]
