91Ó°ÊÓ

The discriminant is a key concept in understanding quadratic equations. For any quadratic equation of the form \(ax^{2} + bx + c = 0\), the discriminant \(D\) is given by the formula:
\[D = b^{2} - 4ac\]

• If \(D > 0\), the quadratic equation has two distinct real solutions.
• If \(D = 0\), the equation has exactly one real solution.
• If \(D < 0\), the equation has no real solutions, meaning the solutions are complex.

This tells us how many real solutions a quadratic equation has without solving it directly. For example, in the equation \(r^{2} + 12r + 36 = 0\), the discriminant is \(0\), so it has one real solution.
Meanwhile, for \(8t^{2} - 11t + 5 = 0\), the discriminant is \(-39\), indicating no real solutions.
The equation \(3v^{2} - 5v - 1 = 0\) has a discriminant of \(37\), so it has two real solutions.

Real solutions refer to the values of the variable that satisfy a quadratic equation and are real numbers.
Real solutions occur when the discriminant \(D\) is zero or positive:

• If \(D = 0\), the quadratic equation has one unique real solution. This is also known as a repeated root.
• If \(D > 0\), the quadratic equation has two distinct real solutions.

For instance, the quadratic equation \(r^{2} + 12r + 36 = 0\) has a discriminant of \(0\), implying it has one real solution, which means the parabola touches the x-axis at one point.
In contrast, the equation \(3v^{2} - 5v - 1 = 0\) has a discriminant of \(37\), which is positive, therefore, it has two real solutions. This means the parabola intersects the x-axis at two distinct points.

The quadratic formula is derived from the process of completing the square and provides a method to find the solutions of any quadratic equation:
\[x = \frac{{-b \pm \sqrt{{b^{2} - 4ac}}}}{{2a}}\]
This formula uses the coefficients \(a\), \(b\), and \(c\) from the quadratic equation \(ax^{2} + bx + c = 0\).
The term under the square root, \(b^{2} - 4ac\), is the discriminant \(D\). Depending on the value of \(D\), the nature of the solutions changes:

• If \(D > 0\), there are two real solutions, obtained by using both the \(+\) and \(-\) in the formula.
• If \(D = 0\), there is one real solution, as the part involving \(\pm\sqrt{D}\) becomes zero.
• If \(D < 0\), there are no real solutions, as the square root of a negative number gives an imaginary number.

For the example \(3v^{2} - 5v - 1 = 0\):

• \(a = 3\)
• \(b = -5\)
• \(c = -1\)

Applying the quadratic formula:
\[v = \frac{{-(-5) \pm \sqrt{{(-5)^{2} - 4(3)(-1)}}}}{{2(3)}} = \frac{{5 \pm \sqrt{{25 + 12}}}}{{6}} = \frac{{5 \pm 6.08}}{{6}}\]
Hence, the two solutions are \(v \approx 1.84\) and \(v \approx -0.17\).