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:

\[3x^2-4x-15=0\]


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

\[\Delta = (-4)^2 – 4(3)(-15) = 16 + 180 = 196 \gt 0\]

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


Now, we need to factorize the [polynomial]…/…/polynomials/). We must find two numbers, \(r_1, r_2\) whose sum \(S = r_1 + r_2\) equals \(b = -4\) and whose product \(P= r_1 \cdot r_2\) equals \(a \cdot c = 3 \cdot -15 = -45\). We can use this simple table to find the numbers that satisfy our constraints.

\begin{array}{rrrr} & r_1 & r_2 & P & S \\ \hline & 5 & – 9 & -45 & -4 \\ & -5 & 9 & -45 & 4\\ \end{array}

  • The numbers \(r_1, r_2\) satisfying the constraint are 5 and -9 (row 1).We can rewrite the polynomial as \(ax^2 + r_{1}x + r_{2}x + c\).

The equation becomes: \[3x^2 -9x + 5x – 15 = 0\]

Factoring common terms, we get:

\begin{align*} 3x(x-3)+5(x-3) & = 0\\[0.6em] (3x+5)(x-3) & = 0 \end{align*}

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

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

The solution to the equation is: \[x = -\frac{5}{3} \quad\quad x = 3\]

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