Quadratic Equations – Exercise 1
Summarize
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 \(x = 3\).