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:

\[2x^2-7x+3=0\]


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:

\[\Delta =(-7)^2 – 4(2)(3) = 49-24 = 25 \gt 0\]

\(\Delta \gt 0 \) 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 \(b = -7\) and whose product \(P= r_1 \cdot r_2\) equals \(a \cdot c = 6\). We can use this simple table to find the numbers that satisfy our constraints.

\begin{array}{rrrr} & r_1 & r_2 & P & S \\ \hline & 1 & 6 & 6 & 7 \\ & -1 & -6 & 6 & -7 \\ & 2 & 3 & 6 & 5 \\ & -2 & -3 & 6 & -5 \\ \end{array}

The numbers \(r_1, r_2\) satisfying the constraint are \(-1\) and \(-6\) (row 2). We can rewrite the polynomial as \(ax^2+r_{1}x+r_{2}x+c\).


The equation becomes: \[2x^2-x-6x+3=0\]

Factoring common terms, we get:

\begin{align*} x(2x-1)-3(2x-1) & = 0 \\[0.6em] (2x-1)(x-3) & = 0 \end{align*}

The solutions are the values of \(x\) for which \(2x-1 = 0\) and \(x-3 = 0\). \[2x-1=0 \to x = \frac{1}{2}\] \[x-3=0 \to x=3\]

The solution to the equation is: \[x = \frac{1}{2} \quad \quad x = 3\]

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