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:

\[(x – 4)^2-9 = 0\]


The term, \((x – 4)^2\), represents the square of a binomial of the form \((a-b)^2\) and can be expanded as \(a^2-2ab+b^2\). Thus, the given equation can be rewritten as:

\[(x^2-8x+16)-9 = 0\]


After performing the necessary calculations, the equation can be written in the following form:

\[x^2-8x+7 = 0\]


We can substitute the coefficients \(a= 1, b=-8, c=7\) into the quadratic formula:

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

We obtain:

\[x_{1,2} = \frac{{-(-8) \pm \sqrt{{(-8)^2 – 4(1)(7)}}}}{{2(1)}}\]


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

\begin{align*} x_{1,2} &= \frac{{8 \pm \sqrt{{64 – 28}}}}{2}\\[0.6em] &= \frac{{8 \pm \sqrt{{36}}}}{2} \end{align*}


Finally, by performing the calculations, we obtain:

\[x_1 = \frac{8 – 6}{2} = \frac{2}{2} = 1\] \[x_2 = \frac{8 + 6}{2} = \frac{14}{2} = 7\]

The solution to the equation is: \[x_1 =1 \quad x_2=7\]


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