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

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SUMMARY

The discussion focuses on simplifying the expression \(2\sqrt{7}\sqrt{8-2\sqrt{7}}\) into \(2(7-\sqrt{7})\). The key steps involve rewriting \(8-2\sqrt{7}\) as \((\sqrt{7}-1)^2\), leading to \(\sqrt{8-2\sqrt{7}} = \sqrt{7}-1\). The alternative method of multiplying the radicals using the rule \(\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}\) also confirms the same result, demonstrating the flexibility in simplification techniques.

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

$$2\sqrt{7}\sqrt{8-2\sqrt{7}}=2\sqrt{7}(\sqrt{7}-1)=2(7-\sqrt{7})$$
 
MarkFL said:
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}?
 
RTCNTC said:
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:

$$2\sqrt{7}\sqrt{8-2\sqrt{7}}=2\sqrt{56-14\sqrt{7}}=2\sqrt{49-14\sqrt{7}+7}=2\sqrt{(7-\sqrt{7})^2}=2(7-\sqrt{7})$$
 
MarkFL said:
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?
 
RTCNTC said:
I get it except for 49 in the radical. Where did 49 come from?

$$56=49+7$$
 
MarkFL said:
$$56=49+7$$

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

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