Ref.: module A

This exercise requires understanding the fundamental concepts of quadratic equations and their solution methods. It involves using the quadratic formula

Solve the quadratic equation:

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


The equation is already reduced to the standard form \(ax^2+bx+c= 0\). We can substitute the coefficients \(a=1, b=-5, c=-14\) into the quadratic formula:

\[x_{1,2} = \frac{{-b \pm \sqrt{{b^2-4ac}}}}{{2a}}\]

We obtain:

\[x_{1,2} = \frac{{-(-5) \pm \sqrt{{(-5)^2-4(1)(-14)}}}}{{2(1)}}\]


In this case, the discriminant \(\Delta\) is \(\geq 0\) so the equation admits two distinct real solutions.

\begin{align*} x_{1,2} &= \frac{{5 \pm \sqrt{{25+56}}}}{2}\\[0.6em] &= \frac{{5 \pm \sqrt{{81}}}}{2}\\ \end{align*}


Finally, by performing the calculations, we obtain:

\[x_1 = \frac{5+ 9}{2} = \frac{14}{2} = 7\] \[ x_2=\frac{5-9}{2} = -\frac{4}{2} = -2\]

The solution to the equation is: \[x =7 \quad \quad x=-2\]


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