- Identify the coefficients \(a = 0.25\), \(b = -8\), and \(c = 0\) from the standard form \(ax^2 + bx + c = 0\).
- Substitute these into the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
- Calculate \(-b = 8\), \(b^2 - 4ac = (-8)^2 - 0\), and \(2a = 2(0.25)\).
- Confirm that the correct representation is \(x = \frac{8 \pm \sqrt{(-8)^2 - (4)(0.25)(0)}}{2(0.25)}\), which corresponds to option 1.
- Final answer: \(\boxed{1}\).
Explanation
- Identify coefficients of the quadratic equation
The given equation is \(0 = 0.25x^2 - 8x\). To solve this quadratic equation, we first write it in the standard form \(ax^2 + bx + c = 0\). Comparing \(0.25x^2 - 8x + 0 = 0\) to the standard form, we identify the coefficients as:
\[a = 0.25, \quad b = -8, \quad c = 0.\]
- Apply the quadratic formula
The quadratic formula, which gives the solutions for any quadratic equation \(ax^2 + bx + c = 0\), is defined as:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]
Now, we substitute the identified coefficients \(a = 0.25\), \(b = -8\), and \(c = 0\) into this formula.
Starting with the numerator, \(-b = -(-8) = 8\).
Next, the discriminant part under the square root, \(b^2 - 4ac\), becomes:
\[(-8)^2 - 4(0.25)(0).\]
Finally, the denominator \(2a\) becomes:
\[2(0.25).\]
Putting these together, the equation for the solutions is:
\[x = \frac{8 \pm \sqrt{(-8)^2 - (4)(0.25)(0)}}{2(0.25)}.\]
- Select the correct option
Comparing our derived expression with the provided options:
- \(x = \frac{8 \pm \sqrt{(-8)^2 - (4)(0.25)(0)}}{2(0.25)}\)
- \(x = \frac{-0.25 \pm \sqrt{(0.25)^2 - (4)(1)(-8)}}{2(1)}\)
- \(x = \frac{-8 \pm \sqrt{(-8)^2 - (4)(0.25)(0)}}{2(0.25)}\)
- \(x = \frac{0.25 \pm \sqrt{(0.25)^2 - (4)(1)(-8)}}{2(1)}\)
Our derived expression matches option 1 exactly. Thus, option 1 is the correct representation.
Examples
Quadratic equations appear frequently in physics when calculating the trajectory of objects. For example, if you throw a ball, its height over time is described by a quadratic equation. Solving for the time \(t\) when the ball hits the ground (height = 0) involves applying the quadratic formula, allowing you to determine how long the ball stays in the air. This fundamental skill is essential for predicting outcomes in sports, engineering, and various physical sciences.