#$&*
MTH 158
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Multiplying Squ. Roots
** **
Section R.8
Page 78
Problem 36.
** **
(Sq.root of X + Sq.root of 5)^2
When you multiply a sq.root by another sq.root does it cancel the sq.root out? Or does this only happen if the numbers inside the root are the same?
@&
The square root of a number is the number you multiply by itself to get the original number.
So sqrt(49) * sqrt(49) would be 49. Since 7 * 7 = 49, it follows that sqrt(49) is 7.
Similarly sqrt(x) * sqrt)x) must be equal to x, since that's what the square root means. And sqrt(5) * sqrt(5) = 5 for the same reason.
However square roots don't cancel each other out. For example sqrt(x) * sqrt(5) is not equal to 5 x.
If you multiply sqrt(x) * sqrt(5) by itself you get
(sqrt(x) * sqrt(5)) * (sqrt(x) * sqrt(5) )
= sqrt(x) * sqrt(x) * sqrt(5) * sqrt(5)
= x * 5, or more grammatically 5x.
Since sqrt(x) * sqrt(5) multiplied by itself gives us 5 x, it mu
#$&*
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.
** **
Section R.8
Page 79
Problem 63.
** **
(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?
@&
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)).