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:

\[3x^2+25x-18 = 0\]


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:

\[\Delta = (25)^2 – 4(3)(-18) = 625 + 216 = 841 \gt 0\]

\(\Delta \gt 0\) 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 \(b = 25\) and whose product \(P = r_1 \cdot r_2\) equals \(a \cdot c = 3 \cdot -18 = -54 \). 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 & – 27 & -54 & -25 \\ & -2 & 27 & -54 & 25\\ \end{array}

The numbers \(r_1, r_2\) satisfying the constraint are \(-2\) and \(27\) (row 1). Then we need to rewrite the polynomial as \(ax^2 + r_{1}x + r_{2}x + c\).


The equation becomes: \[3x^2-2x+27x-18 = 0\]

Factoring common terms, we get:

\begin{align*} x(3x-2)+9(3x-2) &= 0 \\ (3x-2)(x+9) &= 0 \end{align*}

The solutions are the values of \(x\) for which \(3x-2= 0\) and \(x+9 = 0\).

\[3x-2 = 0 \to 3x=2 \to x = \frac{2}{3}\] \[x + 9 = 0 \to x = -9\]

The solution to the equation is:

The solution to the equation is: \[x_1 =\frac{2}{3} \quad \quad x_2=-9\]

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