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Algebra/expansion ( including substitution )

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

2. if a - 1/a =3 ; find :
 (i)a^2 + 1/a^2    (ii)a^3 - 1/a^3

Answer
1. Multiplying through by a gives a² - 4a - 1 = 0.

This means, by quadratic equation, that a = (4ħsqrt(16+4))/2.
That is the same as a = (4ħsqrt(20))/2.

Since sqrt(20) = sqrt(4*5) = 2*sqrt(5), both 4 and that term can be divided by 2.
This gives a = 2 ħ sqrt(5).

Note that this looks like it means that a² = 9 ħ 4*sqrt(5) and a³ = 38 ħ 17*sqrt(5).

Use the '+' term to find one result and the '-' term to find the other.

To simplify the solution, multiply the numerator and denominator of the fraction with the square-root in the denominator by the conjugate of the denominator.  This will make the denominator an integer (no square-roots any more).

2. Multiplying through by a gives a² - 3a - 1 = 0.
The quadratic equation tells us that a = (3ħsqrt(9+4))/2 = (3ħsqrt(13))/2.

Put that in for a in (i) and (ii).
Note that it looks like a² = (22ħ6*root(13))/4 and a³ = (144ħ40*sqrt(13))/8.

Again, use the conjugate of the denominator on the fraction with the square-root in the denominator.

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Scott A Wilson

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Any algebraic question you've got. That includes question that are linear, quadratic, exponential, etc.

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I have solved story problems, linear equations, parabolic equations. I have also solved some 3rd order equations and equations with multiple variables.

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Documents at Boeing in assistance on the manufacturiing floor.

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MS at math OSU in mathematics at OSU, 1986. BS at OSU in mathematical sciences (math, statistics, computer science), 1984.

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Both my BS and MS degrees were given with honors.

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