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

🔗 Related Articles You Might Like:

📰 x = \frac{45}{5} = 9 📰 Thus, there are $\boxed{9}$ thymine bases. 📰 A physicist working on particle collisions finds that the ratio of protons to neutrons in a certain isotopic sample is $2:1$. If there are 8 protons, what is the total number of nucleons (protons and neutrons) in the sample? 📰 For 3 5 📰 For Any Triangle The Radius R Of The Inscribed Circle Is Given By 📰 For Z In 0Circ 360Circ Evaluate 📰 Forget Dog Poopskunk Poop Is The Ultimate No Contraption To Avoid 📰 Forward To Confidence Why Small Boobs Get Overratedand What Works Better 📰 Found A Mysterious Something Blue Borrowednew Old And Bound To Go Viralheres How 📰 Found The Perfect Sorrow Synonymyou Wont Believe How Powerful It Sounds 📰 Found These Southern Green Beanstheir Flavor Is So Good Youll Demand More 📰 Frac0 06 0 0 Rightarrow Texthorizontal Line 📰 Frac1X Frac1Y Frac1Z Frac12 Frac12 Frac12 Frac32 📰 Frac1X Frac1Y Frac1Z Geq Frac32 📰 Frac302424 126 📰 Frac4 03 0 Frac43 📰 Frac63 Geq Frac3Frac1X Frac1Y Frac1Z 📰 Fracddt6T2 4T 30 12T 4 0 Rightarrow T Frac13

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