# Quadratic Equation B5

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https://algebrica.org/exercises/quadratic-equation-b-5/
Fetched from algebrica.org test 4593; source modified 2025-03-06T18:05:22.

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

\\[
x^2 + 8x + 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\\) to ensure the equation admits solutions in the field of real numbers. Substituting the coefficients of the equation into \\(\\Delta\\), we get:

\\[
\\Delta = (8)^2 - 4(1)(15) = 64 - 60 = 4 \\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 = 8\\) and whose product \\(P = r\_1 \\cdot r\_2\\) equals \\(a \\cdot c = 1 \\cdot 15 = 15\\). We can use this simple table to find the numbers that satisfy our constraints.

\\begin{array}{rrrr} & r\_1 & r\_2 & P & S \\\\ \\hline & 3 & 5 & 15 & 8 \\\\ & -3 & -5 & 15 & -8\\\\ \\end{array}

The numbers \\(r\_1, r\_2\\) satisfying the constraint are 3 and 5. Then we need to rewrite the polynomial as \\(ax^2 + r\_{1}x + r\_{2}x + c\\).

* * *

The equation becomes:

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

Factoring common terms, we get:

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

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

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



\\[
x+5= 0 \\to x = -5
\\]

The solution to the equation is:

\\[
x\_1=-3, \\ x\_2=-5
\\]

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}
\\]