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:

\[2x^2 + 10x + 11 = 0\]


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:

\[x_{1,2} = \frac{{-10 \pm \sqrt{{10^2-4(2)(11)}}}}{{2(2)}}\]


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

\begin{align*} x_{1,2} &= \frac{{-10 \pm \sqrt{{100 – 88}}}}{4}\\[0.6em] &= \frac{{-10 \pm \sqrt{{12}}}}{4}\\[0.6em] &= \frac{{-10 \pm \sqrt{{3 \cdot 2^2}}}}{4}\\[0.6em] &= \frac{{-10 \pm 2 \sqrt{{3}}}}{4}\\ \end{align*}


Finally, by performing the calculations, we obtain:

\[ x_{1,2} = \frac{-5 \pm \sqrt{3}}{2}\]

The solution to the equation is: \[x_1=\frac{-5 + \sqrt{3}}{2} \quad x_2=\frac{-5 – \sqrt{3}}{2}\]


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