Source: algebrica.org - CC BY-NC 4.0 https://algebrica.org/exercises/quadratic-equation-a-10/ Fetched from algebrica.org test 3487; source modified 2025-03-06T16:11:23.
Solve the equation:
9x^2 - 5=0
In this case, we have an incomplete quadratic equation, where $b$, the coefficient of the linear term is equal to zero. We can rewrite the equation as:
x^2=\frac{c}{a}
In this case, $a$ the coefficient of the quadratic term $x^2$ and the constant $c$ have different signs, so the equation admits two distinct real solutions:
x_{1,2} = \pm \sqrt{\frac{5}{9}}
Taking the $1/9$ out of the root, we get:
x_{1,2}=\pm \frac{\sqrt{5}}{3}
The solution to the equation is:
x_1 =-\frac{\sqrt{5}}{3} \quad x_2 =\frac{\sqrt{5}}{3}
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}