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

[mathdad]

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

View attachment 7806

 

[Greg]

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

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

 

[topsquark]

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

-Dan

 

[Greg]

Check the last term.

Post corrected. Thanks Dan.

 

[mathdad]

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?