Question
Solving using the Quadratic Formula. \( \begin{array}{l}x^{2}-13 x-48=0 \\ x=3, x=-4 \\ x=-2, x=-4 \\ x=-3, x=16 \\ x=-3, x=-4\end{array} \)
Ask by Burns Watson. in the United States
Nov 06,2024
Real Tutor Solution
Tutor-Verified Answer
Answer
The solutions are \(x=-3\) and \(x=16\).
Solution
Solve the equation \( x^{2}-13x-48=0 \).
Solve the quadratic equation by following steps:
- step0: Solve by factoring:
\(x^{2}-13x-48=0\)
- step1: Factor the expression:
\(\left(x-16\right)\left(x+3\right)=0\)
- step2: Separate into possible cases:
\(\begin{align}&x-16=0\\&x+3=0\end{align}\)
- step3: Solve the equation:
\(\begin{align}&x=16\\&x=-3\end{align}\)
- step4: Rewrite:
\(x_{1}=-3,x_{2}=16\)
The solutions to the quadratic equation \(x^{2}-13x-48=0\) using the Quadratic Formula are \(x=-3\) and \(x=16\).
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Simplify this solution Beyond the Answer
Ready to dive into the quadratic formula fun? The quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), comes to the rescue for solving equations of the form \(ax^2 + bx + c = 0\). With our equation \(x^2 - 13x - 48 = 0\), we plug in \(a = 1\), \(b = -13\), and \(c = -48\) to find the roots. Now, onto real-world application! Quadratic equations aren’t just math class material; they model real-world scenarios like projectile motion. For example, if you throw a ball, its height over time can be represented by a quadratic equation. By solving it, you find when the ball reaches the ground, ensuring you don’t inadvertently throw a little too close to your neighbor's window!
