Source: algebrica.org - CC BY-NC 4.0 https://algebrica.org/exercises/quadratic-equation-a-5/ Fetched from algebrica.org test 3469; source modified 2025-03-06T16:13:23.
Solve the quadratic equation:
2x^2 + 10x + 11 = 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:
x_{1,2} = \frac{{-10 \pm \sqrt{{10^2-4(2)(11)}}}}{{2(2)}}
In this case, the discriminant $\Delta$ is $\geq 0$ so the equation admits two distinct real solutions.
\begin{align*} x_{1,2} &= \frac{{-10 \pm \sqrt{{100 - 88}}}}{4}\\[0.6em] &= \frac{{-10 \pm \sqrt{{12}}}}{4}\\[0.6em] &= \frac{{-10 \pm \sqrt{{3 \cdot 2^2}}}}{4}\\[0.6em] &= \frac{{-10 \pm 2 \sqrt{{3}}}}{4}\\\\ \end{align*}
Finally, by performing the calculations, we obtain:
x_{1,2} = \frac{-5 \pm \sqrt{3}}{2}
The solution to the equation is:
x_1=\frac{-5 + \sqrt{3}}{2} \quad x_2=\frac{-5 - \sqrt{3}}{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}