Quadratic Equation B1

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Source: algebrica.org - CC BY-NC 4.0 https://algebrica.org/exercises/quadratic-equation-b-1/ Fetched from algebrica.org test 4487; source modified 2025-03-06T18:03:26.

Solve the quadratic equations using the factorization method.

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 equation’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}