Can the Algebraic Expression $8-2\sqrt7$ be Simplified to a Perfect Square?

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Discussion Overview

The discussion revolves around whether the algebraic expression $8-2\sqrt7$ can be simplified to a perfect square. Participants explore the mathematical reasoning behind this expression, including the conditions for it to be a perfect square and the implications of such a simplification.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant proposes that $8-2\sqrt7$ can be expressed as $(\sqrt7-1)^2$ by solving the equations derived from the expansion of $(a+b\sqrt7)^2$.
  • Another participant questions the correctness of the last term in the expansion of $(a+b\sqrt7)^2$, prompting a correction from the original poster.
  • A later reply reiterates the claim that $8-2\sqrt7=(\sqrt7-1)^2$ and asks about the algebra involved in taking the square root of both sides.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there are corrections and questions raised about the mathematical steps involved, indicating ongoing debate regarding the simplification of the expression.

Contextual Notes

There are unresolved mathematical steps and potential dependencies on definitions related to perfect squares and algebraic manipulation.

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See picture. Show that the RHS = LHS without using a calculator.

View attachment 7806
 

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Well, let's see...

Is $8-2\sqrt7$ a perfect square number?

$$(a+b\sqrt7)^2=a^2+2ab\sqrt7+7b^2$$

$$a^2+7b^2=8$$

$$2ab=-2$$

$$\implies a=-1,b=1$$

Hence $8-2\sqrt7=(\sqrt7-1)^2$

$$\sqrt7-\sqrt{8-2\sqrt7}=\sqrt7-(\sqrt7-1)=\sqrt7-\sqrt7+1=1$$
 
greg1313 said:
Well, let's see...

Is $8-2\sqrt7$ a perfect square number?

$$(a+b\sqrt7)^2=a^2+2ab\sqrt7+7b$$
Check the last term.
[math](a+b\sqrt7)^2=a^2+2ab\sqrt7+7b^2[/math]

-Dan
 
topsquark said:
Check the last term.

Post corrected. Thanks Dan.
 
greg1313 said:
Well, let's see...

Is $8-2\sqrt7$ a perfect square number?

$$(a+b\sqrt7)^2=a^2+2ab\sqrt7+7b^2$$

$$a^2+7b^2=8$$

$$2ab=-2$$

$$\implies a=-1,b=1$$

Hence $8-2\sqrt7=(\sqrt7-1)^2$

$$\sqrt7-\sqrt{8-2\sqrt7}=\sqrt7-(\sqrt7-1)=\sqrt7-\sqrt7+1=1$$

What would the algebra look like if I take the square root on both sides? Is it tedious?
 

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