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MTH 158

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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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question form

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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.

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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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