+0

# Special Manipulations Help

0
104
7

1. Suppose $$a+\frac{1}{a} =3$$,

a) Find $$a^2 +\frac{1}{a^2}$$

b) Find $$a^4 +\frac{1}{a^4}$$

c) Find $$a^3 + \frac{1}{a^3}$$

2. I'm thinking of two numbers. The sum of my numbers is 14 and the product of my numbers is 46. What is the sum of the squares of my numbers?

3. Simplify $$\sqrt{7-\sqrt{13}} - \sqrt{7+\sqrt{13}}$$

May 22, 2021

#1
+2

First  one

(a)

(a  + 1/a)^2    =   a^2  +  2a (1/a)  + 1/a^2    =    3^2

a^2   + 2  +  1/a^2   =  9

a^2  + 1/a^2   =  7

(b)

a^4  +  1/a^4

(a^2  + 1/a^2)^2    =   a^4  +   2 a^2 (1/a^2)  +  1/a^4     =  49

a^4  + 2  +  1/a^4   =  49

a^4  + 1/a^4    =   47

(c)

a^3   +  1/a^3    =   (a  + 1/a) ( a^2   -  a(1/a)  + 1/a^2)   =

(3)  ( a^2  + 1/a^2   -  1)   =

(3)  ( 7 - 1)  =

18   May 22, 2021
#6
0

Thank you so much

shananigans  May 22, 2021
#2
+1

2.

a + b = 14

ab = 46

(a+b)^2 - 2ab = a^2 + b^2

Take it from here. :))

=^._.^=

May 22, 2021
#7
0

104 right?

Thanks!

shananigans  May 22, 2021
#3
+1

x = sqrt(7 - sqrt(13)) - sqrt(7 + sqrt(13))

x^2 = 7 - sqrt(13) + 7 + sqrt(13) - 2*sqrt(36)

x^2 = 14 - 2*6

x^2 = 2

x = sqrt(2) or - sqrt(2)

However, sqrt(7+sqrt(13)) > sqrt(7-sqrt(13)), so x is a negative number.

x = -sqrt(2)

=^._.^=

May 22, 2021
#4
+1

Very  nice  on  number 3 , catmg.....I wouldn't have  noticed  that  "trick"  !!!!!   CPhill  May 22, 2021
#5
+1

Awww, thank you. :))

=^._.^=

catmg  May 22, 2021