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Algebra/expansion ( including substitution )


""1. if a - 1/a =3 ; find :
 (i)a^2 + 1/a^2    (ii)a^3 - 1/a^3
2. if a - 1/a =4 ; find :
 (i)a^2 + 1/a^2    (ii)a^4 + 1/a^4 (iii)a^3 - 1/a^3"
thankyou !"

1.  If a - 1/a  = 3, then subtracting 3 from both sides and multiplying by a gives
a - 3a - 1 = 0.  This yields a = (3 root(13))/2.
This makes a = (22 + 6*root(13))/4 or a = (22 - 6*root(13))/4.

Multiplying the numerator and denominator by 1/2 gives
a = (11 + 3*root(13))/2 or (11 - 3*root(13))/2.

Note that for the 1st a, a = (238 + 66*root(13))/4 = (118 + 33*root(13))/2 or
a = (238 + 66*root(13))/4 = (118 + 33*root(13))/2.

For the 2nd a, it would be a = (118 - 33*root(13))/2.

Now if it asked for a + 1/a, then that is
(119 + 33*root(13))/2 + 2/(119 + 33*root(13)) and
(119 - 33*root(13))/2 + 2/(119 - 33*root(13)).

Rationalize these two answer by multiplying the 2nd fraction of each by its conjugate on top and bottom, then combine terms on top.  Once this is done, combine the fractions be getting the same value on the denominator.

Multiply the 1st value for a by a to get the 1st value of a and
the 2nd value of a by the 2nd value of a to get a.

2. This problem is solved in a similar fashion.
This time, I believe a = 8 + root(3) and a = 8 - root(3).

For this problem, (i) and (iii) are gotten the same way.
For (ii), a can be squared again or a and a can be multiplied together.


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Scott A Wilson


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