91Ó°ÊÓ

Quadratic equations are mathematical expressions that follow the standard form: \( ax^2 + bx + c = 0 \). In these equations, \( a \), \( b \), and \( c \) are coefficients, with \( a eq 0 \). The presence of \( x^2 \) indicates that the equation is quadratic. Such equations represent parabolas when graphed on a coordinate plane. They are essential in algebra as they describe various natural phenomena and practical applications, from physics to economics.

The equation provided in the exercise, \( 25q^2 - 1 = 0 \), fits this form, with coefficients as follows:

• \( a = 25 \)
• \( b = 0 \)
• \( c = -1 \)

Here, the absence of the \( b \) term simplifies the equation, but it remains quadratic given the power of 2 in \( q^2 \). By understanding the role of each coefficient, solving these equations becomes more intuitive.

Solving quadratic equations can be approached in various ways, including factoring, completing the square, or using the quadratic formula. Each method has its circumstances where it is most efficient. The quadratic formula is especially useful when other methods are cumbersome or impossible.

The quadratic formula is: \[q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula yields the solutions for any quadratic equation. By substituting the coefficients \( a \), \( b \), and \( c \) into the formula, we can solve for the variable \( q \).

In our example, substituting:

• \( a = 25 \)
• \( b = 0 \)
• \( c = -1 \)

Plugging these into the formula provides the steps to the answer. The formula simplifies the process, handling equations where factoring might be too complex or ambiguous.

Quadratic solutions arise from evaluating the quadratic formula, leading to potential solutions as roots of the quadratic equation. The phrase "roots" refers to the values of \( q \) that satisfy the equation, making the quadratic expression equal to zero.

Using the quadratic formula, we first compute the discriminant, \( \sqrt{b^2 - 4ac} \). For our exercise: \[\sqrt{0^2 - 4(25)(-1)} = \sqrt{100}\]This simplifies to \( \pm 10 \).

Consequently, we substitute back into the quadratic formula:

• For \( q_1 = \frac{0 + 10}{50} = \frac{1}{5} \)
• For \( q_2 = \frac{0 - 10}{50} = -\frac{1}{5} \)

These two solutions, \( \frac{1}{5} \) and \( -\frac{1}{5} \), represent the points where the parabola crosses the x-axis. Such solutions reveal the essential nature of quadratic equations and their graphs.