# Quadratic Equation B1

*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 equations:

\[x^2-5x + 6 = 0\]

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 \(\geq0\) to ensure the equation admits solutions in the field of real numbers. Substituting the coefficients of the equation into \(\Delta\), we get:

\[\Delta =(-5)^2-4(1)(6) = 1 \gt 0\]

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

To find the equationâ€™s solutions using the factorization method, we must find two numbers whose sum equals \(b\), or \(-5\) in this case, and whose product is equal to \(a \cdot c\), or \(6\). It is immediate to see that the polynomial \(x^2-5x + 6\) is factorizable as the product of two binomials \((x-2)\) and \((x-3)\).

We obtain:

\[x^2 – 5x + 6 = (x-2)(x-3) = 0\]

The values of \(x\) that make the product null of \((x-2)(x-3)\) are \(x=2 \) and \(x=3\).

The solution to the equation is:

\[x_1= 2, \ x_2 =3\]

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