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

February 22, 2026

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

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.

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.

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

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

Both sides match—solution verified. Similar verification holds for the second solution.

Step 5: Real-World Applications

Quadratic equations model many real-life phenomena, such as:

Projectile motion: distance over time often follows a parabolic path.
Profit maximization: revenue minus cost functions typically produce quadratic expressions.
Physics: energy and motion relationships can be represented by quadratics.

Understanding how to solve equations like u² + 1 = 3u helps analyze and predict outcomes in science, engineering, and economics by modeling constraints and relationships algebraically.

Step 6: Exploring the Discriminant

The discriminant (Δ = b² - 4ac) tells us about the nature of the solutions: Δ = 9 − 4(1)(1) = 5 > 0 → two distinct real solutions Δ = 0 → one real solution Δ < 0 → complex (non-real) solutions

In this case, Δ = 5 confirms two real, irrational roots—consistent with irrational numbers arising from √5.

Conclusion

Rearranging u² + 1 = 3u into u² - 3u + 1 = 0 transforms an expression into a solvable quadratic form. Using the quadratic formula allows precise calculation of the two real solutions: (3 + √5)/2 and (3 − √5)/2, each verified and meaningful in both theoretical and applied contexts. Mastering these techniques equips learners with powerful tools for tackling complex equations across mathematics and science.

Keywords: quadratic equation, solve quadratic, u² - 3u + 1 = 0, quadratic formula, discriminant, real solutions, irrational numbers, algebra, mathematics tutorial

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