MHB How to Simplify a Radicand by Breaking it into Two Parts?

[mathdad]

How does (2sqrt{7})(sqrt{8 - 2sqrt{7}) become
2(7 - sqrt{7})?

 

[MarkFL]

Well, we may write:

$$8-2\sqrt{7}=7-2\sqrt{7}+1=(\sqrt{7}-1)^2$$

And so:

$$\sqrt{8-2\sqrt{7}}=\sqrt{(\sqrt{7}-1)^2}=\sqrt{7}-1$$

Thus:

$$2\sqrt{7}\sqrt{8-2\sqrt{7}}=2\sqrt{7}(\sqrt{7}-1)=2(7-\sqrt{7})$$

 

[mathdad]

Well, we may write:

$$8-2\sqrt{7}=7-2\sqrt{7}+1=(\sqrt{7}-1)^2$$

And so:

$$\sqrt{8-2\sqrt{7}}=\sqrt{(\sqrt{7}-1)^2}=\sqrt{7}-1$$

Thus:

$$2\sqrt{7}\sqrt{8-2\sqrt{7}}=2\sqrt{7}(\sqrt{7}-1)=2(7-\sqrt{7})$$

Nicely done as always. What if I decided to multiply the two given radicals using the rule sqrt{a}*sqrt{b} = sqrt{ab}?

 

[MarkFL]

Nicely done as always. What if I decided to multiply the two given radicals using the rule sqrt{a}*sqrt{b} = sqrt{ab}?

Well if you did that, you would have:

$$2\sqrt{7}\sqrt{8-2\sqrt{7}}=2\sqrt{56-14\sqrt{7}}=2\sqrt{49-14\sqrt{7}+7}=2\sqrt{(7-\sqrt{7})^2}=2(7-\sqrt{7})$$

 

[mathdad]

Well if you did that, you would have:

$$2\sqrt{7}\sqrt{8-2\sqrt{7}}=2\sqrt{56-14\sqrt{7}}=2\sqrt{49-14\sqrt{7}+7}=2\sqrt{(7-\sqrt{7})^2}=2(7-\sqrt{7})$$

I get it except for 49 in the radical. Where did 49 come from?

 

[MarkFL]

I get it except for 49 in the radical. Where did 49 come from?

$$56=49+7$$

 

[mathdad]

$$56=49+7$$

I understand now. You broke 56 into two parts. This is a fine "trick" or step to simplify the radicand.