More Square Roots: (sqrt3 - sqrt5)(sqrt3 - sqrt5), etc.

alwaysalillost

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Joined
May 15, 2007
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4
Find the square root of the following:

1. 3 sqrt2 + sqrt50

2.(sqrt 2 - 3)(sqrt2 + 3)

3. (sqrt3 - sqrt5)(sqrt3 - sqrt5)

Please help me solve the following problems. Thanks so much!
 
Hi alwaysalillost! Hopefully I'll find you in this explanation. :)

1. \(\displaystyle \L \;3\sqrt{2}\,+\,\sqrt{50}\)

32\displaystyle 3\sqrt{2} is already simplified. But 50\displaystyle \sqrt{50} can be written as \(\displaystyle \sqrt{5^2\,\cdot\,2\)

Pull out the pair of fivess in the square root: 5052\displaystyle \sqrt{50}\,\to\,5\sqrt{2}

So since we have the same number in the square root for both terms, we add the number in front of the same square roots and keep the same square root:

\(\displaystyle \L \;3\sqrt{2}\,+\,5\sqrt{2}\,=\,(5\,+\,3)\sqrt{2}\,=\,8\sqrt{2}\)

2. \(\displaystyle \L \;(\sqrt{2}\,-\,3)(\sqrt{2}\,+\,3)\)

This is a difference of squares. That means:   (ab)(a+b)=a2b2\displaystyle \;(a\,-\,b)(a\,+\,b)\,=\,a^2 - b^2

3. \(\displaystyle \L \;(\sqrt{3}\,-\,sqrt{5})(\sqrt{3}\,-\,sqrt{5})\)

You have the same binomial. So: (a+b)(a+b)=a2+2ab+b2\displaystyle (a\,+\,b)(a\,+\,b)\,=\,a^2\,+\,2ab\,+\,b^2
 
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