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MTH 158
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** Question Form_labelMessages **
Multiplying Squ. Roots
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Section R.8
Page 78
Problem 36.
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(Sq.root of X + Sq.root of 5)^2
When you multiply a sq.root by another sq.root does it cancel the sq.root out? Or does this only happen if the numbers inside the root are the same?
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The square root of a number is the number you multiply by itself to get the original number.
So sqrt(49) * sqrt(49) would be 49. Since 7 * 7 = 49, it follows that sqrt(49) is 7.
Similarly sqrt(x) * sqrt)x) must be equal to x, since that's what the square root means. And sqrt(5) * sqrt(5) = 5 for the same reason.
However square roots don't cancel each other out. For example sqrt(x) * sqrt(5) is not equal to 5 x.
If you multiply sqrt(x) * sqrt(5) by itself you get
(sqrt(x) * sqrt(5)) * (sqrt(x) * sqrt(5) )
= sqrt(x) * sqrt(x) * sqrt(5) * sqrt(5)
= x * 5, or more grammatically 5x.
Since sqrt(x) * sqrt(5) multiplied by itself gives us 5 x, it must be the case that sqrt(x) * sqrt(5) is the same as sqrt( 5 x).
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To calculate the result you are asked for in this problem:
(sqrt(x) + sqrt(5))^2
= (sqrt(x) + sqrt(5) ) * (sqrt(x) + sqrt(5) )
= sqrt(x) (sqrt(x) + sqrt(5) ) + sqrt(5) (sqrt(x) + sqrt(5) )
= sqrt(x) sqrt(x) + sqrt(x) sqrt(5) + sqrt(5) sqrt(x) + sqrt(5) sqrt(5)
= x + 2 sqrt(x) sqrt(5) + 5
= x + 5 + 2 sqrt(5x).
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#$&*
mth 158
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Sq.root as denom.
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Section R.8
Page 79
Problem 63.
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(9 / 8) ^(3/2)
So: (Sq.root of 9^3) / (Sq.root of 8^3)
Therefore the top would equal 27.
Leaving the bottom as sq.root of 64 * sq.root of 4 * sq.root of 2
The answer to this problem is 27 * Sq.root of 2 / 32
So how does the Sq.root of 4 lose it's sq.root and still equal 4?
Also, how does the Sq.root of 2 end up on the top?
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sqrt(8^3) = sqrt(8*2 * 8)
= sqrt(8^2) * sqrt(8)
= sqrt(8^2) * sqrt(4 * 2)
= sqrt(8^2) * sqrt(4) * sqrt(2)
= 8 * 2 * sqrt(2).
So the result would be
27 / (8 * 2 * sqrt(2))
= 27 / (16 sqrt(2)).
However simplified form doesn't have a radical in the denominator.
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27 / (16 sqrt(2))
needs to be expressed without the radical in the denominator.
This is easily done. We simply multiply numerator and denominator by sqrt(2), obtaining
( 27 / (16 sqrt(2)) ) * ( sqrt(2) / sqrt(2) )
27 sqrt(2) / ( (16 sqrt(2)) * sqrt(2) )
= 27 sqrt(2) / (16 * 2)
= 27 sqrt(2) / 32.
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