Ref.: module A

This exercise requires understanding the fundamental concepts of quadratic equations and their solution methods. It involves using the quadratic formula

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. \[\nexists \hspace{10px} x \in \mathbb{R}\]