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Topic 9.2 (Rational Exponents)
Every positiveinteger has a positive and negative square root.
The square rootof: 16 = 4
The square rootsign: is called the root symbol.
16 is calledthe radicand
The number inthe v of the radical is called the index number or root number. If
there isnt any number show it is 2, otherwise if you have a cube root& etc.
the3 will be written in the index (root) number position.
A square root of a negative numbercan never be a real number.
REMEMBER: The square root of 16 is 4, why , because 4 squaredis 16.
Therefore: the square root of -16 cannot be found in the real number system,
because 4 or -4 squared will not be 16.
25 = 5 The primary or principal root which is alwayspositive.
- 25 = -5 The secondary root which is alwaysnegative.
25 = 5 The double root which is always positive andnegative.
If the index (root) number is even, a negative number will not result in a real
If the index (root) number is odd, a negative number will work, giving a
negative answer. - 8 = - 2
The above reads the cube root of -8 is -2.
am/n = ( a)m = (am)This says the nth root of a to themth power.
The nth root of a.
82/3 = ( 8)2 = (82) This says the cube root of 8 squared.
The cube root of 8.
When solving this kind of problem change the original to its corresponding
REMEMBER:The numerator of the fraction is the exponent of the radicand and
the denominator is the index (root) number.
It is easier to change the fractional exponent to a root raised to a
power. This means is easier to take the root and than raise it to
82/3 = ( 8)2 = (2)2 = 4
The cube root of 8 = 2
165/4 = ( 16)5 = (2)5 =32
The fourth root of 16 = 2.
x-n = 1/xn and 1/x-n = xn This means when the exponentis negative in
the numerator (upstairs), allterms affected by
the negative exponents are moved to the
denominator (downstairs) and the exponents
are changed to a positive exponent
Also when the terms have a negative
exponent in the denominator they are moved
to the numerator and the exponent is
changed to a positive exponent.
Example: (All answers must have apositive exponent.)
2x-2y3/ w-1z-3 Move the x term to the denominator, and the w term to
the numerator and change the sign of the exponents.
2wy3 / x2z3 Ans.
(2-1x4y-3/ z-2)-2 Move the terms with the negative exponents up or down
(x4z2 / 2y3)-2 and change the exponent to positive. (terms inside the
(2y3 / x4z2)2 Next turn the whole fraction upside down and change the
sign of the outside exponent to positive.
Now raise a power to a power.
4y6 / x8z4 Ans.
There is also a way to work theseproblems by raising a power to a power right at
the start and them changing to apositive exponent. The only problem with this
is you might miss writing down anegative at some point and the problem will be
To simplify a radical expression
Make sure all perfect roots are taken out.
No fractions are left under the radical sign.
The denominator of a fraction doesnt contain a radical.
The index number and the exponent of the radicand do not
contain a common factor.
(72) = (36 2) = 62
(8x4y12z5) = (4 2x4y12z4z) = 2x2y6z2 (2z)
(242x7y3z15) = (121 2x6xy2yz14z) = 11x3yz7 (2xyz)
(1/5) change to 1/5 , multiply the numerator and the
denominator by 5.
5/ (25) = 5 / 5.
(3/2) = 3 / 2 multiply the numerator and the denominator by 2.
6 / 4 = 6 / 2.
A short cut: Multiply the numerator by the denominator and put this in
in the numerator of the answer under the radical and put the old
denominator in the denominator of the answer without the radical.
(3/5) = (15) / 5
5/ 7) multiply the numerator and the denominator by 7.
5 7 / 49 = 5 7/ 7
(6/ 5) multiply the numerator and the denominator by 5.
(30)/ (25) = (30) / 5.
(32) = 3 divide the exponent by the index no. 2
(74) = 72 divide the exponent by the index no. 2.
ADDING AND SUBTRACTING RADICALS
To add or subtract radicals you musthave the same radicand and the
22 + 52 - 32 = 42.
8 + (18) - (50) Simplify allterms (remove all perfect
(4x 2) + (9x 2) - (25x 2)
22 + 32 - 52 = 0
In a problem like: 5 /(2 + 7) You must multiply the numerator and the
denominator by 2 - 7.
This is called the conjugate. Theconjugate is the same two numbers with the sign
Suchas: 5 - 3 theconjugate is 5 + 3.
2 + 5 the conjugate is 2 - 5.
So in the problemabove 5 (2- 7 = 10 - 57 =
(2 + 7)(2 - 7) 4 - 7
10 - 57 = -10 + 57