- Rearrange the equation \(9u^2 + 25 = 30u\) into the standard form \(9u^2 - 30u + 25 = 0\).
- Factor the quadratic as a perfect square trinomial \((3u - 5)^2 = 0\).
- Solve the linear equation \(3u - 5 = 0\) to find the value of \(u\).
- The final answer is \(u = \boxed{\frac{5}{3}}\).
Explanation
- Analysis and Rearrangement of the Quadratic Equation
We are given the quadratic equation \(9u^2 + 25 = 30u\). Our goal is to solve for the variable \(u\). To do this, we first rearrange the equation into the standard quadratic form \(au^2 + bu + c = 0\) by subtracting \(30u\) from both sides:
\[9u^2 - 30u + 25 = 0\]
- Factoring the Quadratic Equation
We look at the standard form \(9u^2 - 30u + 25 = 0\). We notice that this expression is a perfect square trinomial. A perfect square trinomial of the form \(A^2x^2 - 2ABx + B^2\) factors into \((Ax - B)^2\). Here, we have:
\((3u)^2 - 2 \cdot (3u) \cdot 5 + 5^2 = 0\)
This matches the structure, so we can factor the equation as:
\[(3u - 5)^2 = 0\]
- Solving for u
With the equation factored as \((3u - 5)^2 = 0\), we can solve for \(u\) by taking the square root of both sides, which gives:
\[3u - 5 = 0\]
Adding 5 to both sides results in:
\[3u = 5\]
Finally, dividing by 3, we get:
\[u = \frac{5}{3}\]
Thus, the real solution to the equation is \(u = \frac{5}{3}\).
Examples
Quadratic equations frequently appear in real-life physics and engineering problems. For example, if you throw a ball upwards, its height over time is modeled by a quadratic equation. If you want to know when the ball reaches a specific height, you set up the equation and solve for time \(t\). Understanding how to factor these equations efficiently is a fundamental skill in solving for variables in various scientific and practical contexts.