# Quadratic Equation B4
Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/exercises/quadratic-equation-b-4/
Fetched from algebrica.org test 4589; source modified 2025-03-06T18:04:55.
Solve the [quadratic equations](../quadratic-equations.md) using the [factorization method](../factoring-quadratic-equations.md).
\\[
x^2-5x-14 = 0
\\]
* * *
The equations is in the standard form \\(ax^2+bx+c=0\\). First, it is essential to verify the its [discriminant](../quadratic-formula.md) \\(\\Delta = b^2 - 4ac\\) is \\(\\geq0 \\quad\\) to ensure the equation admits solutions in the field of real numbers. Substituting the coefficients of the equation into \\(\\Delta\\), we get:
\\[
\\Delta = (-5)^2-4(1)(-14) = 25 + 56 = 81 \\gt 0
\\]
\\(\\Delta \\gt 0 \\) means the equation has real solutions.
* * *
Now, we need to factorize the polynomial. We must find two numbers, \\(r\_1, r\_2\\) whose sum \\(S = r\_1 + r\_2\\) equals \\(b = -5\\) and whose product \\(P = r\_1 \\cdot r\_2\\) equals \\(a \\cdot c = 1 \\cdot -14 = -14\\). 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 & - 7 & -14 & -5 \\\\ & -2 & 7 & -14 & 5\\\\ \\end{array}
The numbers \\(r\_1, r\_2\\) satisfying the constraint are 2 and -7 (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-7x-14 = 0
\\]
* * *
The equation becomes:
\\[
3x^2 -9x + 5x-15 = 0
\\]
Factoring common terms, we get:
\\begin{align\*} x(x+2)-7(x+2) &= 0 \\[0.6em] (x+2)(x-7) & = 0 \\end{align\*}
The solutions are the values of \\(x\\) for which \\(x+2 = 0\\) and \\(x-7 = 0\\).
\\[
x+2 = 0 \\to x=-2
\\]
\\[
x-7 = 0 \\to x=7
\\]
The solution to the equation is:
\\[
x\_1=-2, \\ x\_2=7
\\]
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}
\\]
