# Quadratic Equation B2
Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/exercises/quadratic-equation-b-2/
Fetched from algebrica.org test 4578; source modified 2025-03-06T18:03:58.
Solve the [quadratic equations](../quadratic-equations.md) using the [factorization method](../factoring-quadratic-equations.md).
\\[
2x^2-7x+3=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 =(-7)^2 - 4(2)(3) = 49-24 = 25 \\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 = -7\\) and whose product \\(P= r\_1 \\cdot r\_2\\) equals \\(a \\cdot c = 6\\). We can use this simple table to find the numbers that satisfy our constraints.
\\begin{array}{rrrr} & r\_1 & r\_2 & P & S \\\\ \\hline & 1 & 6 & 6 & 7 \\\\ & -1 & -6 & 6 & -7 \\\\ & 2 & 3 & 6 & 5 \\\\ & -2 & -3 & 6 & -5 \\\\ \\end{array}
The numbers \\(r\_1, r\_2\\) satisfying the constraint are \\(-1\\) and \\(-6\\) (row 2). We can rewrite the polynomial as \\(ax^2+r\_{1}x+r\_{2}x+c\\).
* * *
The equation becomes:
\\[
2x^2-x-6x+3=0
\\]
Factoring common terms, we get:
\\begin{align\*} x(2x-1)-3(2x-1) & = 0 \\[0.6em] (2x-1)(x-3) & = 0 \\end{align\*}
The solutions are the values of \\(x\\) for which \\(2x-1 = 0\\) and \\(x-3 = 0\\).
\\[
2x-1=0 \\to x = \\frac{1}{2}
\\]
\\[
x-3=0 \\to x=3
\\]
The solution to the equation is:
\\[
x = \\frac{1}{2} \\quad \\quad x = 3
\\]
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}
\\]
