-4(root of 5*5*3) + 3(root of 3**7*7)
-20√(3) + 21√3
which is equal to √3
2007-06-05 04:31:46
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answer #1
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answered by Uncle Under 5
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First you need to factor the numbers under the square-root signs:
75 = 3* 25 = 3*5*5
147 = 3 * 49 = 3*7*7
Notice that both terms have a factor which is a square: 75 has 25, and 147 has 49. We can move the square root of that number outside the radical:
-4(5)sqrt(3) + 3(7)sqrt(3)
-20sqrt(3)+21sqrt(3)
Notice that the two terms have a common factor of sqrt(3), so we can factor that out:
sqrt(3)(-20+21)
sqrt(3)(1)
sqrt(3)
And there's your answer. I hope this helps. Good luck on other problems of this type.
2007-06-05 04:34:11
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answer #2
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answered by Anonymous
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= (- 4).(5).√3 + (3).(7).√3
= - 20.√3 + 21√3
= √3
2007-06-05 04:54:13
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answer #3
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answered by Como 7
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no calculators needed here.
-4sqrt(75) + 3sqrt(147)
-4sqrt(25)sqrt(3) + 3sqrt(49)sqrt(3)
-20sqrt(3) + 21sqrt(3)
sqrt(3)
and that's it. one can add squareroots as long as the number inside is the same. and one can break apart sqrt roots into two roots, especially usefull if one of the roots is a perfect square, such as 25.
2007-06-05 04:32:08
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answer #4
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answered by Anonymous
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-4*sqrt(75) + 3* sqrt(147)
= -4 * sqrt(5*5*3) + 3 * sqrt(7*7*3)
= -4 * 5 * sqrt(3) + 3 * 7 * sqrt(3)
= -20 * sqrt(3) + 21 * sqrt(3)
= (-20 + 21) * sqrt(3)
= sqrt(3)
2007-06-05 04:32:11
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answer #5
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answered by to0pid 2
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+root 3
2007-06-05 04:32:02
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answer #6
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answered by shreyas 2
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you guys are a bunch of dorks.
2007-06-05 04:44:35
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answer #7
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answered by Mr. Ed 2
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root3
2007-06-05 04:36:55
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answer #8
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answered by kannu p 1
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