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:

\[(4x + 8)\left(\frac{1}{2}x – 6\right) = 0\]


We can use the distributive property to expand the equation and obtain the following:

\begin{align*} \frac{4}{2}x^2-24x + \frac{8}{2}x-48 = 0 \\[0.6em] 2x^2-24x +4x-48 = 0 \\[1em] 2x^2-20x-48 = 0 \end{align*}


The coefficients \(a, b\) and \(c\) have 2 as common multiplier. We can factor out the number and obtain:

\[x^2-10x-24 = 0\]


The equation is now reduced to the standard form \(ax^2+bx+c= 0\). We can substitute the coefficients \(a=2, b=20, c=48\) into the quadratic formula:

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

We obtain:

\[x_{1,2}= \frac{{-(-10) \pm \sqrt{{(-10)^2-4(1)(-24)}}}}{{2(1)}}\]


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

\begin{align*} x_{1,2} &= \frac{{10 \pm \sqrt{{100 + 96}}}}{2}\\[1em] x_{1,2} &= \frac{{10 \pm \sqrt{{196}}}}{2} \end{align*}


Finally, by performing the calculations, we obtain:

\[x_1 = \frac{{10 + 14}}{2} = 12\] \[x_2 = \frac{{10 – 14}}{2} = -2\]

The solution to the equation is: \[x =12 \quad \quad x =-2\]


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