Question #4824d

1 Answer
Aug 18, 2017

#x^2 + 16/x^2 = 28#

Explanation:

The first thing to notice here is that

#x^2 + 16/x^2#

can be written as

#(x + 4/x)^2 - 8#

This is the case because

#(x + 4/x)^2 = x^2 + 2 * color(red)(cancel(color(black)(x))) * 4/color(red)(cancel(color(black)(x))) + (4/x)^2#

# = x^2 + 8 + 16/x^2#

So if you subtract #8# from both sides of this equation, you will end up with

#(x + 4/x)^2 - 8 = x^2 + color(red)(cancel(color(black)(8))) + 16/x^2 - color(red)(cancel(color(black)(8)))#

which is

#(x + 4/x)^2-8 = x^2 + 16/x^2" "color(blue)("(*)")#

Now, you know that

#x = 3 - sqrt(5)#

This means that you have

#x + 4/x = 3 - sqrt(5) + 4/(3 - sqrt(5))#

If you rationalize the denominator of #4/(3 - sqrt(5))# by multiplying both the numerator and the denominator by #3 + sqrt(5)#, the conjugate of #3 - sqrt(5)#, you will end up with

#3 - sqrt(5) + (4 * (3 + sqrt(5)))/((3 - sqrt(5))(3 + sqrt(5))#

At this point, you can use the fact that

#color(blue)(ul(color(black)(a^2 - b^2 = (a-b)(a + b))))#

to say that

#(3 - sqrt(5))(3 + sqrt(5)) = 3^2 - (sqrt(5))^2#

# = 9 - 5#

# = 4#

This means that you have

#3 - sqrt(5) + (color(red)(cancel(color(black)(4))) * (3 + sqrt(5)))/color(red)(cancel(color(black)(4))) = 3 - color(red)(cancel(color(black)(sqrt(5)))) + 3 + color(red)(cancel(color(black)(sqrt(5))))#

# = 6#

Therefore, you can say that

#x + 4/x = 6#

Plug this back into equation #color(blue)("(*)")# to get

#(x + 4/x)^2 - 8 = 6^2 - 8 = 28#

This means that you have

#x^2 + 16/x^2 = 28#