- Rearrange the equation to standard form: \(x^2 + x - 5 = 0\).
- Identify the coefficients \(a = 1, b = 1, c = -5\).
- Apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the solutions.
- The resulting solutions are \(\frac{-1 \pm \sqrt{21}}{2}\), which corresponds to option d.
- Final answer: \(\boxed{d}\)
Explanation
- Standard Form and Coefficients
The problem asks to solve the quadratic equation \(x^2 = 5 - x\) using the quadratic formula. First, we must write the equation in the standard quadratic form, which is \(ax^2 + bx + c = 0\). Adding \(x\) and subtracting \(5\) from both sides of the original equation yields:
\[x^2 + x - 5 = 0\]
Here, the coefficients are \(a = 1\), \(b = 1\), and \(c = -5\).
- Applying the Quadratic Formula
The quadratic formula is given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Substituting the identified coefficients \(a=1, b=1, c=-5\) into the formula, we have:
\[x = \frac{-(1) \pm \sqrt{(1)^2 - 4(1)(-5)}}{2(1)}\]
Simplifying the expression under the square root, known as the discriminant:
\[x = \frac{-1 \pm \sqrt{1 - (-20)}}{2}\]
\[x = \frac{-1 \pm \sqrt{1 + 20}}{2}\]
\[x = \frac{-1 \pm \sqrt{21}}{2}\]
Thus, the solutions to the equation are \(\frac{-1 \pm \sqrt{21}}{2}\). Comparing this result with the given options, we identify that it matches option d.
Examples
Quadratic equations are fundamental in modeling real-world phenomena where acceleration or area is involved. For instance, if you were launching a model rocket and wanted to determine the time \(t\) at which it reaches a certain height relative to the ground, you would use a quadratic equation where the squared term represents gravity. Solving for \(t\) helps you predict when the rocket will reach its target altitude or return to the ground, allowing for precise control and safety.