Understanding the Equation: u² - 2u + 1 + 2u - 2 + 2 = u² + 1

When solving algebraic expressions, simplifying both sides is key to verifying mathematical identities or solving equations. One such expression often examined for pattern recognition and identity confirmation is:

Left Side:
u² - 2u + 1 + 2u - 2 + 2

Understanding the Context

Right Side:
u² + 1

But does this equation actually hold true? Let’s break it down step-by-step to understand the validity, simplify both sides, and clarify its meaning.

Simplifying the Left Side

Solved -alpha(2u_2n - u_2n+1 - u_2n-1) = m_1 d^2u_2n/dt^2 | Chegg.com

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Solved -alpha(2u_2n - u_2n+1 - u_2n-1) = m_1 d^2u_2n/dt^2 | Chegg.com
1 8π u 1 u 2 (u 2 1 − 1)(u 2 2 − 1)e − u 2 1 2 e − u 2 2 2 | Download ...
Exemple pour f = u 1 + u 2 + u −2 1 u −3 2 , g = u 1 u 2 et h = u −2 1 ...
Solved U1+U2={u1+u2:u1∈U1 and u2∈U2} Show that U1+U2 is a | Chegg.com
what is value u1′=−4u1−2u2+cost+4 sint | Chegg.com

Key Insights

We begin with the full left-hand side expression:
u² - 2u + 1 + 2u - 2 + 2

Group like terms intelligently:

  • Quadratic term: u²
  • Linear terms: -2u + 2u = 0u (they cancel out)
  • Constant terms: 1 - 2 + 2 = 1

So, the simplified left side becomes:
u² + 1

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Final Thoughts

Comparing to the Right Side

The simplified left side u² + 1 matches exactly with the right side:
u² + 1

✅ Therefore, the equation holds:
u² - 2u + 1 + 2u - 2 + 2 = u² + 1 is a true identity for all real values of u.

Why This Identity Matters

While the equation is algebraically correct, its deeper value lies in demonstrating how canceling terms—especially linear terms—can reveal hidden equivalences. This is especially useful in:

  • Algebraic proofs
  • Simplifying complex expressions
  • Identifying perfect square trinomials (this expression rearranges to (u – 1)²)

Indeed, recognizing the original left side as a grouping that eliminates the linear –2u + 2u term highlights the power of reordering and combining terms.

Final Summary