# Quadratic Equation B8
Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/exercises/quadratic-equation-b-8/
Fetched from algebrica.org test 5007; source modified 2025-03-06T18:07:04.
Solve the [quadratic equations](../quadratic-equations.md) using the [factorization method](../factoring-quadratic-equations.md).
\\[
3x^2+25x-18 = 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 = (25)^2 - 4(3)(-18) = 625 + 216 = 841 \\gt 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 = 25\\) and whose product \\(P = r\_1 \\cdot r\_2\\) equals \\(a \\cdot c = 3 \\cdot -18 = -54 \\). 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 & - 27 & -54 & -25 \\\\ & -2 & 27 & -54 & 25\\\\ \\end{array}
The numbers \\(r\_1, r\_2\\) satisfying the constraint are \\(-2\\) and \\(27\\) (row 1). Then we need to rewrite the polynomial as \\(ax^2 + r\_{1}x + r\_{2}x + c\\).
* * *
The equation becomes:
\\[
3x^2-2x+27x-18 = 0
\\]
Factoring common terms, we get:
\\begin{align\*} x(3x-2)+9(3x-2) &= 0 \\\\ (3x-2)(x+9) &= 0 \\end{align\*}
The solutions are the values of \\(x\\) for which \\(3x-2= 0\\) and \\(x+9 = 0\\).
\\[
3x-2 = 0 \\to 3x=2 \\to x = \\frac{2}{3}
\\]
\\[
x + 9 = 0 \\to x = -9
\\]
The solution to the equation is:
The solution to the equation is:
\\[
x\_1 =\\frac{2}{3} \\quad \\quad x\_2=-9
\\]
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}
\\]
