Ref.: module B

This exercise requires understanding the fundamental concepts of quadratic equation and their solution methods. It involves using the factorization method.

Solve the quadratic equation:

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


Expanding the equation using the distributive property, we get:

\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 simplify the equation which is now brought into the standard form of a quadratic equation:

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


The equations is in the standard form \(ax^2+bx+c=0\). It is essential to verify the its discriminant \(\Delta = b^2 – 4ac\) is \(\geq0\) to ensure the equation admits solutions in the field of real numbers. Substituting the coefficients of the equation into \(\Delta\), we get:

\[\Delta = (-10)^2 – 4(1)(-26) = 100 + 96 = 196 \geq 0\]

\(\Delta \gt 0\) means the equation has real solutions.


Now, we need to factorize the polynomial. We must find two numbers, \(r_1, r_2\) whose sum \(S = r_1 + r_2\) equals \(b = -10\) and whose product \(P = r_1 \cdot r_2\) equals \(a \cdot c = 1 \cdot -24 = -24 \). We can use this simple table to find the numbers that satisfy our constraints.

\begin{array}{rrrr} & r_1 & r_2 & P & S \\ \hline & 2 & – 12 & -24 & -10 \\ & -2 & +12 & -24 & 10\\ \end{array}

The numbers \(r_1, r_2\) satisfying the constraint are \(2\) and \(-12\) (row 1). Then we need to rewrite the polynomial as \(ax^2 + r_{1}x + r_{2}x + c\).


The equation becomes: \[x^2+2x-12x-24 = 0\]

Factoring common terms, we get:

\begin{align*} &x(x+2)-12(x+2) = 0 \\[0.6em] &(x+2)(x-12) = 0 \end{align*}

The solutions are the values of \(x\) for which \(x-2= 0\) and \(x-12 = 0\).

\[x -2 = 0 \to x=-2\] \[x -12 = 0 \to x=12\]

The solution to the equation is:

The solution to the equation is: \[x_=-2 \quad \quad x_1 =12\]

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