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: \[x^2 = 5x-6\]


First, we need to rewrite the second-degree equation in its standard form \(ax^2+bx+c = 0\). By collecting all terms on the left side of the equal sign, we obtain::

\[x^2-5x+6 = 0\]


After reducing the equation to its standard form, we can substitute the coefficients \(a=1, b=-5, c=6\) into the quadratic formula:

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

We obtain:

\[x_{1,2}= \frac{{-(-5) \pm \sqrt{{(-5)^2-4(1)(6)}}}}{{2(1)}}\]


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 – 24}}}}{2}\\ x_{1,2} &= \frac{{5 \pm 1 }}{2} \end{align*}


Finally, by performing the calculations, we obtain:

\[x_1 = \frac{5 – 1}{2} = \frac{4}{2} = 2\] \[x_2 = \frac{5 + 1}{2} = \frac{6}{2} = 3\]

The solution to the equation is: \[x_1=2 \quad\quad x_2=3\]


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