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:

\[12x^2 +17x – 5 = 0\]


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:

\[\Delta = (17)^2 – 4(12)(-5) = 529 \geq 0\]

\(\Delta \gt 0\) 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 \(b = 17\) and whose product \(P = r_1 \cdot r_2\) equals \(a \cdot c = 12 \cdot -5= -60\). We can use this simple table to find the numbers that satisfy our constraints. \(-60\) can be factored into 20 and – 3 and that the sum of the two numbers gives 17.


The equations becomes:

\[12x^2 +20x -3x – 5 = 0\]

Factoring the common terms, we get: \begin{align*} & 12x^2 -3x +20x- 5 = 0 \\[0.6em] & 3x(4x-1) + 5(4x-1) = 0 \\[0.6em] & (3x + 5)(4x-1) = 0 \end{align*}

The solutions are the values of \(x\) for which \(3x + 5 = 0\) and \(4x-1 = 0\)

\[ 3x + 5 = 0 \to 3x=-5 \to x = -\frac{5}{3} \] \[4x-1 = 0 \to 4x=1 \to x = \frac{1}{4}\]

The solution to the equation is:

\[x = -\frac{5}{3} \quad \quad x = \frac{1}{4}\]

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