the abterms from the numerator and
denominator, leaving ab.- c.First, cube the 4g^2 term. Cube the constant 4
and multiply the exponent ofgby 3: (4g^2 )^3 =
64 g^6. Next, multiply 64g^6 by g^4. Add the expo-
nents of the gterms. (64g^6 )(g^4 ) = 64g.^10 Finally,
taking the square root of 64g^10 yields 8g^5 , since
(8g^5 )(8g^5 ) = 64g.^10 - e.First, find the square root of 9pr. 9 pr=
9 pr= 3pr.The denominator (pr) has a
negative exponent, so it can be rewritten in the
numerator with a positive exponent. The
expression prcan be written as (pr)
since a value raised to the exponent ^12 is
another way of representing the square root of
the value. The expression is now 3(pr)(pr).
To multiply the prterms, add the exponents.^12
+ ^32 = ^42 = 2, so =3(pr)(pr)= 3(pr)^2 = 3p^2 r^2.- e.Substitute 20 forn:(^220 5 ) =
(10 5 ) = (10 5 ). Cancel the 5
terms and multiply the fraction by 10:
(10 5 ) = 5(1 2 0)= ^520 = 25- c.^ = = =
- d. =^4 = ^481 = ^43 ^4 = 3
- d.x^2 + 4x+ 4= (x+ 2)^2 = x+ 2
- d.^4 32 x^8 = ^42 ^4 2 (x^2 )^4 = ^42 ^4 ^42 ^4 (x^2 )^4
= 2x2 4√ 4 - b.^4 x^21 = ^4 (x^5 )^4 x= ^4 (x^5 )^4 ^4 x= x^5 ^4 x
- b.^3 54 x^5 = ^32 33 x^3 x^2 = 3x^32 x^2
- a.x^3 + 40x^2 + 40 0 x= x(x^2 + 40 x+ 400)=
x(x+ 20)^2 = (x+ 20) x
Set 38 (Page 93)- b.–25= 25 (–1)= 5 ^2 i^2 =
5 ^2 i^2 = 5i - a.–32= (32(–1)= 32 –1= (4√2)(i) =
4 i 2 - a.
- 48 + 2 27 – 75 = – 4 ^2 3 + 2 3 ^2 3
- 5 ^2 3 = –4 3 + 6 3 – 5 3 =
(–4 + 6 – 5) 3 = –3 3 - d. 3 3 + 4 5 – 8 3 = (3 – 8) 3 + 4 5 =
–5 3 + 4 5 - d.First, simplify each radical expression. Then,
because the variable/radical parts are alike, we
can add the coefficients:
xy 8 xy^2 + 3y^2 18 x^3 =xy(2y) 2 x+ 3y^2 (3x)
2 x= 2xy^2 2 x+ 9xy^2 2 x= 11xy^2 2 x
Same Same- c.We first simplify each fraction. Then, we
find the LCD and add.
+ = + = + =
^33 + 4^ ^55 = =
- a.(5 – 3 )(7 + 3 ) = 5(7) + 5( 3 ) – 7( 3 )
- 3 ^2 = 35 + (5 – 7) 3 – 3 = 32 – 2 3 - b.(4 + 6 )(6 – 15 ) = 24 – 4 15 + 6 6 –
90 = 24 –4 15 + 6 6 – 3 10 - a. = = =
= –2 +i- c.(4 + 2i)(4 –2i) = 16 – (2i)^2 = 16 – 2^2 i^2 =
16 – (4)(–1) = 16 + 4= 20 - d.(4 + 2i)^2 = 16 +(4)(2i) + (2i)(4) + (2i)^2 =
16 + 16i+ 2^2 i^2 = 16 +16i–4 = 12 + 16i
5(–2 + i)
5–10 + 5i
5
–10 + ^25 –1
5
–10 + –25
5^29 ^2
15
^9 ^2 + 20^2
15
^2
3
^3 ^2
5^4 ^2
3
^3 ^2
5
^32
9
^18
25
^32
9
^18
25^243
3
^4243
^43 ^5 ^5
3
^5 ^2 ^5
3 ^2
^125
9
^125
9^5
2 5 ^5
2 5
^25
20 20 + 5
20 ^32 ^12^32 ^12^12^32ANSWERS & EXPLANATIONS–} }