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

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

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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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