Source: algebrica.org - CC BY-NC 4.0 https://algebrica.org/exercises/quadratic-equation-a-4/ Fetched from algebrica.org test 3464; source modified 2025-03-06T16:13:44.
Solve the quadratic equation:
x^2-5x-14 = 0
The equation is already reduced to the standard form $ax^2+bx+c= 0$. We can substitute the coefficients $a=1, b=-5, c=-14$ 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)(-14)}}}}{{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+56}}}}{2}\\[0.6em] &= \frac{{5 \pm \sqrt{{81}}}}{2}\\\\ \end{align*}
Finally, by performing the calculations, we obtain:
x_1 = \frac{5+ 9}{2} = \frac{14}{2} = 7
x_2=\frac{5-9}{2} = -\frac{4}{2} = -2
The solution to the equation is:
x =7 \quad \quad x=-2
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}