Math question about powers

Question

A number is one more than its reciprocal. Find the sum of the fourth power of the number and the fourth power of the reciprocal.

Answer 1

Given the parameters of the problem, where a number is one more than its reciprocal, we can come up with this equation:

 

x = (1/x) + 1

 

We can then multiply both sides of the equation by x and continue on like this:

 

x² = (x/x) + x

 

x² = 1 + x

 

x² - x - 1 = 0

 

Now we have a quadractic equation, and we can plug in the values into the quadratic formula:

 

x = -b ± √(b² - 4ac)

               2a

 

x = (1) ± √((1) - 4 (1)(-1)

                  2(1)

 

x = 1.618, -.618

 

If we add the numbers in the second sentence of your problem together like this:

 

x⁴ + (1/x)⁴

 

where x can be either of the two numbers we found from the quadratic formula, the answer will be 7.

Answer 2

Assume a number is a/b.

 

Then its reciprocal is b/a.

 

Thus, a/b

= (b/a) + 1

= (bb/ab) + (ab/ab)

= [(bb + ab)/ab] <-- common denominator and add

 

Therefore the sum is (a/b)^4 + [(bb+ab)/ab]^4

 

This should be simplified down to one fraction... unfortunately, it is too had to type.  Just remember to find common denominators and combine like terms.