Quadratic Equations – Exercise 1

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Exercise

Solve the quadratic equation:

x^2 = 5x-6

First, we need to rewrite the second-degree equation in its standard form $ax^2+bx+c = 0$. By collecting all terms on the left side of the equal sign, we obtain::

x^2-5x+6 = 0

After reducing the equation to its standard form, we can substitute the coefficients $a=1, b=-5, c=6$ into the quadratic formula:

x_{1,2} = \frac{{-b \pm \sqrt{{b^2-4ac}}}}{{2a}}

We obtain:

x_{1,2}= \frac{{-(-5) \pm \sqrt{{(-5)^2-4(1)(6)}}}}{{2(1)}}

In this case, the discriminant $\Delta$ is $\geq 0$ so the equation admits two distinct real solutions. \begin{align*} x_{1,2} &= \frac{{5 \pm \sqrt{{25 - 24}}}}{2}\\ x_{1,2} &= \frac{{5 \pm 1 }}{2} \end{align*}

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}

Finally, by performing the calculations, we obtain:

x_1 = \frac{5 - 1}{2} = \frac{4}{2} = 2
x_2 = \frac{5 + 1}{2} = \frac{6}{2} = 3

The solution to the equation is:

x_1=2 \quad\quad x_2=3

Factorization as an alternative approach

A more immediate approach can be used to solve the proposed equation, taking advantage of the fact that this is a relatively simple trinomial that can often be factored at a glance with a bit of practice. Starting from its standard form, we can solve the quadratic equation 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 $(x-3)$.

In situations where the correct pair of numbers is not as evident, it is helpful to organize the possible combinations in a small table listing the candidate values $r_1$ and $r_2$, together with their product $P$ and sum $S$. This allows us to quickly identify the pair that satisfies both conditions:

\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}

Only the last row satisfies both conditions $P = 6$ and $S = -5$. We obtain:

x^2 - 5x + 6 = (x-2)(x-3) = 0

The values of $x$ that make the product $(x - 2)(x - 3) = 0$ are $x = 2$ and $x = 3$.

These are exactly the same solutions obtained through the quadratic formula, confirming that both methods, factorization and the general formula, lead to the same results when the trinomial is easily factorable.