- Rearrange the quadratic equation to standard form: \(x^2 - 7x + 12 = 0\).
- Factor the quadratic trinomial into: \((x - 3)(x - 4) = 0\).
- Apply the zero product property to set each factor to zero: \(x - 3 = 0\) or \(x - 4 = 0\).
- Conclude that the solutions are \(x=3\) and \(x=4\), which is option (a).
- The final answer is \(\boxed{a}\).
Explanation
- Rewriting the Equation in Standard Form
To solve the quadratic equation \(x^2 + 12 = 7x\), we first rewrite it in standard form by moving all terms to one side. Subtracting \(7x\) from both sides, we obtain:
\[x^2 - 7x + 12 = 0\]
This is a standard quadratic equation in the form \(ax^2 + bx + c = 0\), where \(a=1, b=-7,\) and \(c=12\).
- Factoring the Quadratic Equation
To solve the equation \(x^2 - 7x + 12 = 0\), we look to factor the quadratic expression \(x^2 - 7x + 12\). We need two numbers that multiply to \(12\) and add up to \(-7\). By inspection, these numbers are \(-3\) and \(-4\). Thus, we can write the quadratic as:
\[(x - 3)(x - 4) = 0\]
- Applying the Zero Product Property and Finding Solutions
The zero product property states that if the product of two factors is zero, then at least one of the factors must be zero. This gives us two linear equations:
\(x - 3 = 0\) or \(x - 4 = 0\)
Solving these equations individually:
- For \(x - 3 = 0\), we add 3 to both sides to get \(x = 3\).
- For \(x - 4 = 0\), we add 4 to both sides to get \(x = 4\).
Thus, the solutions to the original equation are \(x = 3\) and \(x = 4\), which corresponds to option (a).
Examples
The zero product property is incredibly useful in real-world scenarios where you need to find critical points or solve for unknowns. For example, if you are modeling the flight path of a projectile where the height is given by a quadratic expression, such as \(h(t) = -t^2 + 7t - 12\), you can find the times at which the object hits the ground (height = 0) by factoring the equation \(0 = -(t-3)(t-4)\). This allows you to quickly determine that the projectile hits the ground at \(t=3\) and \(t=4\) seconds. Understanding how to factor these equations helps in fields ranging from engineering and physics to economics.