# Quadratic Equation B3

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https://algebrica.org/exercises/quadratic-equation-b-3/
Fetched from algebrica.org test 4583; source modified 2025-03-06T18:04:27.

Solve the [quadratic equations](../quadratic-equations.md) using the [factorization method](../factoring-quadratic-equations.md).

\\[
3x^2-4x-15=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 = (-4)^2 - 4(3)(-15) = 16 + 180 = 196 \\gt 0
\\]

\\(\\Delta \\gt 0 \\) means the equation has real solutions.

* * *

Now, we need to [factorize](../factoring-quadratic-equations.md) the \[polynomial\]…/…/polynomials/). We must find two numbers, \\(r\_1, r\_2\\) whose sum \\(S = r\_1 + r\_2\\) equals \\(b = -4\\) and whose product \\(P= r\_1 \\cdot r\_2\\) equals \\(a \\cdot c = 3 \\cdot -15 = -45\\). We can use this simple table to find the numbers that satisfy our constraints.

\\begin{array}{rrrr} & r\_1 & r\_2 & P & S \\\\ \\hline & 5 & - 9 & -45 & -4 \\\\ & -5 & 9 & -45 & 4\\\\ \\end{array}

-   The numbers \\(r\_1, r\_2\\) satisfying the constraint are 5 and -9 (row 1).We can rewrite the polynomial as \\(ax^2 + r\_{1}x + r\_{2}x + c\\).

* * *

The equation becomes:

\\[
3x^2 -9x + 5x - 15 = 0
\\]

Factoring common terms, we get:

\\begin{align\*} 3x(x-3)+5(x-3) & = 0\\[0.6em] (3x+5)(x-3) & = 0 \\end{align\*}

The solutions are the values of \\(x\\) for which \\(3x+5 = 0\\) and \\(x-3 = 0\\).

\\[
3x+5 = 0 \\to 3x=-5 \\to x = -\\frac{5}{3}
\\]



\\[
x-3 = 0 \\to x=3
\\]

The solution to the equation is:

\\[
x = -\\frac{5}{3} \\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}
\\]