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simplify (3 - sqrt[2]) / (4 + 2sqrt[2])
- Thread starter humakhan
- Start date
No, sorry, not correct. You can't cancel like that.
Multiply top and bottom by the conjugate of the denominator:
\(\displaystyle \L\\\frac{(3-\sqrt{2})}{(4+2\sqrt{2})}\cdot\frac{(4-2\sqrt{2})}{(4-2\sqrt{2})}\)
Expand out the numerator and denominator, then simplify. Don't cancel the conjugate. You need it to simplify.
Multiply top and bottom by the conjugate of the denominator:
\(\displaystyle \L\\\frac{(3-\sqrt{2})}{(4+2\sqrt{2})}\cdot\frac{(4-2\sqrt{2})}{(4-2\sqrt{2})}\)
Expand out the numerator and denominator, then simplify. Don't cancel the conjugate. You need it to simplify.
No, humakhan. You seem to be having trouble with the basics. You can't multiply that way.
You found the dot product, I suppose, if I had to call it something.
\(\displaystyle \L\\(3-\sqrt{2})(4-2\sqrt{2})=12-10\sqrt{2}+4=16-10\sqrt{2}\)
The denominator you have correct only because the middle terms cancel.
\(\displaystyle \L\\\frac{16-10\sqrt{2}}{8}=\frac{8-5\sqrt{2}}{4}\)
You found the dot product, I suppose, if I had to call it something.
\(\displaystyle \L\\(3-\sqrt{2})(4-2\sqrt{2})=12-10\sqrt{2}+4=16-10\sqrt{2}\)
The denominator you have correct only because the middle terms cancel.
\(\displaystyle \L\\\frac{16-10\sqrt{2}}{8}=\frac{8-5\sqrt{2}}{4}\)
3 - sqrt(2) * 4 - 2sqrt(2) = (3 - sqrt(2)) (4 - 2sqrt(2))
-------------- ---------------- ---------------------------------
4 + 2sqrt(2) * 4 - 2sqrt(2) = (4 + 2sqrt(2)) (4 - 2sqrt(2)
equals
12 - 4sqrt(2) - 6 sqrt(2) + 2 * 2 = 16 - 10sqrt(2)
------------------------------------- ------------------
16 + 8sqrt(2) - 8sqrt(2) - 4 * 2 = 8
equals
2 - 1.25 sqrt(2)
-------------- ---------------- ---------------------------------
4 + 2sqrt(2) * 4 - 2sqrt(2) = (4 + 2sqrt(2)) (4 - 2sqrt(2)
equals
12 - 4sqrt(2) - 6 sqrt(2) + 2 * 2 = 16 - 10sqrt(2)
------------------------------------- ------------------
16 + 8sqrt(2) - 8sqrt(2) - 4 * 2 = 8
equals
2 - 1.25 sqrt(2)
Before you try problems of this difficulty you need to know that you can only cancel from numerator to denominator if the things you are canceling are FACTORS of the entire numerator and the entire denominator. For example you CANNOT cancel in this illustration:
3 + ( x-y)
-------------
3 - (x-y)
You can cancel here:
3(x-y)
---------
6(x-y)
3 + ( x-y)
-------------
3 - (x-y)
You can cancel here:
3(x-y)
---------
6(x-y)