Ref.: module B

This exercise requires understanding the fundamental concepts of quadratic equation and their solution methods. It involves using the factorization method.

Solve the quadratic equation:

\[x^2 + 8x + 15 = 0\]


The equations is in the standard form \(ax^2+bx+c=0\). First, it is essential to verify the its discriminant \(\Delta = b^2 – 4ac\) 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 the polynomial. 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 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. \[\nexists \hspace{10px} x \in \mathbb{R}\]