Solving the Quadratic Equation: u² + 1 = 3u and Its Transformed Form

Understanding how to solve quadratic equations is a fundamental skill in algebra, essential for students and math enthusiasts alike. One common transformation in quadratic problems is rearranging expressions like u² + 1 = 3u into standard quadratic form u² - 3u + 1 = 0. This not only simplifies solving but also reveals important properties about the equation’s solutions. In this article, we’ll explore how to manipulate the equation, solve it using the quadratic formula, interpret the solutions, and apply these techniques in real-world scenarios.

Understanding the Context

Step 1: Rearranging the Equation

The original equation is:
u² + 1 = 3u

To solve for u, bring all terms to one side to form a standard quadratic equation:
u² - 3u + 1 = 0

This transformation is key because it allows direct application of quadratic solving methods.

u^2 + 1 = 3u \Rightarrow u^2 - 3u + 1 = 0

Key Insights

Step 2: Identifying Coefficients

A standard quadratic equation is written as:
au² + bu + c = 0

Comparing this with our equation:

  • a = 1
  • b = -3
  • c = 1

Continue Reading

1.08⁶ = 1.3605 × 1.1664 ≈ 1.5869
Rounded to nearest tenth: 23.8 cm
A = 300 × e^(-0.04×6) = 300 × e^(-0.24)

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Final Thoughts

Step 3: Solving Using the Quadratic Formula

The quadratic formula solves for u when the equation is in standard form:
u = [ -b ± √(b² - 4ac) ] / (2a)

Substituting a = 1, b = -3, c = 1:
u = [ -(-3) ± √((-3)² - 4(1)(1)) ] / (2 × 1)
u = [ 3 ± √(9 - 4) ] / 2
u = [ 3 ± √5 ] / 2

Thus, the two solutions are:
u = (3 + √5)/2 and u = (3 − √5)/2

Step 4: Verifying Solutions

It’s always wise to verify solutions by substituting back into the original equation. Let’s check one:
Let u = (3 + √5)/2

Compute u²:
u = (3 + √5)/2 → u² = [(3 + √5)/2]² = (9 + 6√5 + 5)/4 = (14 + 6√5)/4 = (7 + 3√5)/2

Now, left side:
u² + 1 = (7 + 3√5)/2 + 1 = (7 + 3√5 + 2)/2 = (9 + 3√5)/2

Right side:
3u = 3 × (3 + √5)/2 = (9 + 3√5)/2