Solve Algebra: 16+(y-6)^2=9+(y-5)^2

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Discussion Overview

The discussion revolves around the algebraic manipulation of the equation 16+(y-6)^2=9+(y-5)^2, specifically focusing on the implications of taking square roots and the conditions under which certain simplifications are valid. The scope includes mathematical reasoning and clarification of algebraic principles.

Discussion Character

  • Mathematical reasoning, Conceptual clarification

Main Points Raised

  • One participant questions why the constants 16 and 9 are not square rooted when simplifying the equation involving square roots.
  • Another participant asserts that if \sqrt{a} = \sqrt{b}, then a must equal b, prompting a discussion on the implications of this for the given expressions.
  • A participant expresses confusion about the simplification process, indicating a misunderstanding of how square roots apply to the expressions involved.
  • It is noted that the expression \sqrt{16 + (y-6)^2} does not simplify to 16+(y-6)^2 without additional context or manipulation.
  • One participant suggests that squaring both sides of the equation can eliminate the square roots, leading to a different form of the equation.

Areas of Agreement / Disagreement

Participants demonstrate a mix of understanding and confusion regarding the manipulation of square roots and the conditions under which simplifications can be made. There is no consensus on the best approach to the problem, and some participants express differing levels of clarity on the topic.

Contextual Notes

There are unresolved assumptions regarding the conditions under which square roots can be simplified and the implications of squaring both sides of an equation. The discussion does not fully address potential restrictions or limitations in the algebraic transformations discussed.

DB
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When

[tex]\sqrt{16 + (y-6)^2}=\sqrt{9+(y-5)^2}[/tex]

is simplified to:

[tex]16+(y-6)^2=9+(y-5)^2[/tex]

How come the 16 and 9 arent square rooted?

Thanks
 
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Do you agree that [itex]\sqrt{a} = \sqrt{b}[/itex] implies a = b?

If so, what happens when you set a = 16 + (y - 6)^2 and b = 9 + (y - 5)^2?
 
ahhh, o you its because they are equal, i got mixed up thanks.
But [tex]\sqrt{16 + (y-6)^2}[/tex]
doesn't simplify to [tex]16+(y-6)^2[/tex] right?
 
Correct. It would have if the expression was [itex]\sqrt{ (16 + (y - 6)^2)^2 }[/itex] though ;)
 
ya thanks muzza
 
If you square both sides of the equation,
[tex]\sqrt{16- (y-6)^2}= \sqrt{9- (y-5)^2}[/tex]
you eliminate the two square roots and you get
16- (y-6)2= 9- (y-5)2.
 

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