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:

\[7x^2+x+5=0\]


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{{-(1) \pm \sqrt{{(1)^2 – 4(7)(5)}}}}{{2(7)}}\\[0.8em] &= \frac{{-1 \pm \sqrt{{1 – 140)}}}}{{14}}\\[0.8em] &= \frac{{-1 \pm \sqrt{{-139}}}}{{14}}\\ \end{align*}


In this case, the discriminant \(\Delta\) is \(\leq 0\) which means the equation has no real solutions.

The solution to the equation is: \[\nexists \hspace{10px} x \in \mathbb{R}\]


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