Source: algebrica.org - CC BY-NC 4.0 https://algebrica.org/exercises/quadratic-equation-a9/ Fetched from algebrica.org test 3484; source modified 2025-03-06T16:11:49.
Solve the quadratic equation:
7x^2+x+5=0
The equation is already reduced to the standard form $ax^2+bx+c= 0$. We can substitute the coefficients $a=2, b=10, c=11$ into the quadratic formula:
x_{1,2} = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}
We obtain:
\begin{align*} x_{1,2} &= \frac{{-(1) \pm \sqrt{{(1)^2 - 4(7)(5)}}}}{{2(7)}}\\[0.8em] &= \frac{{-1 \pm \sqrt{{1 - 140)}}}}{{14}}\\[0.8em] &= \frac{{-1 \pm \sqrt{{-139}}}}{{14}}\\\\ \end{align*}
In this case, the discriminant $\Delta$ is $\leq 0$ which means the equation has no real solutions.
The solution to the equation is:
\nexists \hspace{10px} x \in \mathbb{R}
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}