# Quadratic Equation A2

*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 quadratic equation:

\[\frac{1}{2}x^2 + \frac{\sqrt{3}}{2}x + \frac{5}{8}=0\]

The equation is already reduced to the standard form \(ax^2+bx+c= 0\). The coefficients \(a\), \(b\) and \(c\) have as a common multiplier \(1/2\). The equation becomes:

\[ \frac{1}{2} \cdot \left( x^2 +\sqrt{3}x+\frac{5}{4} \right) = 0 \]

Multiplying both sides of the equation by 2, we get: \[x^2+\sqrt{3}x+\frac{5}{4}=0\]

The next step involves substituting the coefficients into \(a, b, c\) the quadratic formula to solve the equation. This process necessitates careful consideration of the variables and their respective values. Once the coefficients are inserted, the formula for the variable in question can be solved.

\[x_{1,2} = \frac{{-b \pm \sqrt{{b^2-4ac}}}}{{2a}}\]

- We have:

\begin{align*} x_{1,2} &= \frac{{-(\sqrt{3}) \pm \sqrt{{(\sqrt{3})^2-4(1)(\frac{5}{4})}}}}{{2}}\\[1em] &= \frac{{-\sqrt{3} \pm \sqrt{{3-20}}}}{{2}}\\[1em] &= \frac{{-\sqrt{3} \pm \sqrt{{-17}}}}{{2}}\\ \end{align*}

- In this case, the discriminant \(\Delta = b^2-4ac\) is \(-17< 0\) so 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. \[\nexists \hspace{10px} x \in \mathbb{R}\]