What do you think it is?
Let's find out if you are right:
Since 9 squared is 81, 9 is the square root of 81.
Note that we are only interested in the principal root and since 81
is positive and there is not a sign in front of the radical, our answer
is positive 9. If there had been a negative in front of the radical
our answer would have been -9.
What do you think it is?
Let's find out if you are right:
Since -3 cubed is -27, -3 is the cube root of -27.
Since the root number and the exponent inside are equal and are the even number 2, we need to put an absolute value around x for our answer.
The reason for the absolute value is that we do not know if x is positive or negative. So if we put x as
our answer and it was negative, it would not be a true statement.
What do you think it is?
Let's find out if you are right:
Since there is no such real number that when we square it we get -4,
then the answer is not a real number.
Since we cannot take the square root of 14 and 14 does not have any
factors we can take the square root of, this is as simplified as it gets.
*The cube root of -1 is -1 and the cube root
of 8 is 2
*Simplify the fraction
*The square root of 9 a squared is 3|a|
Can you think of what that factor is?
Let's see what we get when we simplify the first radical:
*Square root of 4 is 2
Square roots are nice to work with in this type of problem because if the radicand is not a perfect square to begin with, we just have to multiply it by itself and then we have a perfect square.
So in this case we can accomplish this by multiplying top and bottom
by the square root of 7:
*Den. now has a perfect square under sq. root
AND
So what would the conjugate of our denominator be?
It looks like the conjugate is .
*Use distributive prop. to multiply the numerators
*In general, product of conjugates is
AND
*Square root of 2 squared is 2 and the square
root of 3 squared is 3
*Divide BOTH terms of num. by -1
Last revised on Dec. 6, 2009 by Kim Seward.
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