System of Equation Challenge (a+b)(b+c)=-1

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Discussion Overview

The discussion revolves around a system of equations involving real numbers \(a\), \(b\), and \(c\). Participants explore the implications of the equations and attempt to find the value of \((a-c)^2+(a^2-c^2)^2\). The scope includes mathematical reasoning and problem-solving related to algebraic identities and equations.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Homework-related

Main Points Raised

  • Participants present a system of equations and seek to derive a specific expression involving \(a\), \(b\), and \(c\).
  • One participant provides a detailed algebraic manipulation to arrive at a potential solution, suggesting that \((a-c)^2+(a^2-c^2)^2 = 160\) under certain conditions.
  • Another participant expresses appreciation for the insights shared and notes that the problem is unsolved on another forum, indicating ongoing interest and exploration of the topic.
  • Repetitive posts indicate a focus on the same algebraic manipulation, with slight variations in presentation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the final value of \((a-c)^2+(a^2-c^2)^2\), as the discussion includes multiple presentations of the same calculations without resolving the underlying assumptions or conditions.

Contextual Notes

The discussion does not clarify the assumptions behind the algebraic manipulations or the conditions under which the derived expression holds true. There is also no resolution on the validity of the calculations presented.

anemone
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Given that $$a,\,b$$ and $$c$$ are real numbers that satisfy the system of equations below:

$$(a+b)(b+c)=-1\\(a-b)^2+(a^2-b^2)^2=85\\(b-c)^2+(b^2-c^2)^2=75$$

Find $$(a-c)^2+(a^2-c^2)^2$$.
 
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anemone said:
Given that $$a,\,b$$ and $$c$$ are real numbers that satisfy the system of equations below:

$$(a+b)(b+c)=-1\\(a-b)^2+(a^2-b^2)^2=85\\(b-c)^2+(b^2-c^2)^2=75$$

Find $$(a-c)^2+(a^2-c^2)^2$$.
This took me much longer than it should have done!
[sp]$$\begin{aligned}(a-c)^2+(a^2-c^2)^2 &= \bigl((a-b) + (b-c)\bigr)^2 + \bigl((a^2-b^2) + (b^2-c^2)\bigr)^2 \\ &= (a-b)^2 + 2(a-b)(b-c) + (b-c)^2 + (a^2-b^2)^2 + 2(a^2-b^2)(b^2-c^2) + (b^2-c^2)^2 \\ &= (a-b)^2 + (a^2-b^2)^2 + (b-c)^2 + (b^2-c^2)^2 + 2(a-b)(b-c) + 2(a^2-b^2)(b^2-c^2) \\ &= 85 + 75 + 2(a-b)(b-c)\bigl(1 + (a+b)(b+c)\bigr) \\ &= 160 + 2(a-b)(b-c)(1 - 1) = 160 \end{aligned}$$

[/sp]
 
Awesome, as always, Opalg!(Yes)

This actually is an unsolved problem at AOPS forum, and I will share the link of this thread at their site now, and thanks again Opalg for your great insight to attack the problem!(Cool)
 
Opalg said:
This took me much longer than it should have done!
[sp]$$\begin{aligned}(a-c)^2+(a^2-c^2)^2 &= \bigl((a-b) + (b-c)\bigr)^2 + \bigl((a^2-b^2) + (b^2-c^2)\bigr)^2 \\ &= (a-b)^2 + 2(a-b)(b-c) + (b-c)^2 + (a^2-b^2)^2 + 2(a^2-b^2)(b^2-c^2) + (b^2-c^2)^2 \\ &= (a-b)^2 + (a^2-b^2)^2 + (b-c)^2 + (b^2-c^2)^2 + 2(a-b)(b-c) + 2(a^2-b^2)(b^2-c^2) \\ &= 85 + 75 + 2(a-b)(b-c)\bigl(1 + (a+b)(b+c)\bigr) \\ &= 160 + 2(a-b)(b-c)(1 - 1) = 160 \end{aligned}$$

[/sp]
Brilliant!

-Dan
 

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