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:

\[x^2-5x-14 = 0\]


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 \(\geq0 \quad\) to ensure the equation admits solutions in the field of real numbers. Substituting the coefficients of the equation into \(\Delta\), we get:

\[\Delta = (-5)^2-4(1)(-14) = 25 + 56 = 81 \gt 0\]

\(\Delta \gt 0 \) 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 \(b = -5\) and whose product \(P = r_1 \cdot r_2\) equals \(a \cdot c = 1 \cdot -14 = -14\). 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 & – 7 & -14 & -5 \\ & -2 & 7 & -14 & 5\\ \end{array}

The numbers \(r_1, r_2\) satisfying the constraint are 2 and -7 (row 1). Then we need to rewrite the polynomial as \(ax^2 + r_{1}x + r_{2}x + c\). The equation becomes: \[x^2+2x-7x-14 = 0\]


The equation becomes: \[3x^2 -9x + 5x-15 = 0\]

Factoring common terms, we get:

\begin{align*} x(x+2)-7(x+2) &= 0 \\[0.6em] (x+2)(x-7) & = 0 \end{align*}

The solutions are the values of \(x\) for which \(x+2 = 0\) and \(x-7 = 0\).

\[x+2 = 0 \to x=-2\] \[x-7 = 0 \to x=7\]

The solution to the equation is: \[x_1=-2, \ x_2=7\]

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. \[\nexists \hspace{10px} x \in \mathbb{R}\]