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:

\[x^2 + 0.4x-0.16 = 0\]


To solve the equation, we can get rid of the decimals by multiplying every coefficient by 100, obtaining:

\[100x^2 + 40x-16 = 0\]


The equation is already reduced to the standard form \(ax^2+bx+c= 0\). We can substitute the coefficients \(a=2, b=10, c=11\) into the quadratic formula:

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

We obtain:

\begin{align*} x_{1,2} &= \frac{-40 \pm \sqrt{40^2-4(100)(-16)}}{2(100)}\\[0.8em] &= \frac{-40 \pm \sqrt{1600+6400}}{200} \end{align*}


In this case, the discriminant \(\Delta\) is \(\geq 0\) so the equation admits two distinct real solutions. by performing the calculations we have:

\begin{align*} x_{1,2} &= \frac{{-40\pm \sqrt{{8000}}}}{200}\\[0.8em] &= \frac{{-40\pm \sqrt{{1600 \times 5}}}}{200}\\[0.8em] &= \frac{{-40\pm \sqrt{{40^2 \times 5}}}}{200}\\[0.8em] &= \frac{{-40\pm 40\sqrt{{5}}}}{200}\\[0.8em] &= \frac{{-1 \pm \sqrt{{5}}}}{5} \end{align*}


Finally, we obtain:

\[x_1 = \frac{{-1 + \sqrt{{5}}}}{5}\] \[x_2= \frac{{-1-\sqrt{{5}}}}{5}\]

The solution to the equation is: \[x = \frac{{-1+\sqrt{{5}}}}{5} \quad x = \frac{{-1-\sqrt{{5}}}}{5}\]


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