Question 1201936
<font color=black size=3>
Answer: <font color=red size=4>Choice (E)  sqrt(21)</font>




Explanation:



Square both sides of the given equation
p+ (1/p) = 5
[ p+ (1/p) ]^2 = 5^2
[ p+ (1/p) ][ p+ (1/p) ] = 25
p*p + p*(1/p) + (1/p)*p + (1/p)(1/p) = 25 .... FOIL rule
p^2 + 1 + 1 + (1/p)^2 = 25
p^2 + 2 + (1/p)^2 = 25
p^2 + (1/p)^2 = 23
I'll refer to this equation as eq2 for a <font color=blue>substitution</font> step later on.


Let q = p - (1/p)
Square both sides to see what happens


p - (1/p) = q
[ p - (1/p) ]^2 = q^2
p*p + p*(-1/p) + (-1/p)*p + (-1/p)(-1/p) = q^2
p^2 - 1 - 1 + (1/p)^2 = q^2
p^2 - 2 + (1/p)^2 = q^2
p^2 + (1/p)^2 - 2 = q^2
[ p^2 + (1/p)^2 ] - 2 = q^2
[ <font color=blue size=4>p^2 + (1/p)^2</font> ] - 2 = q^2
[ <font color=blue size=4>23</font> ] - 2 = q^2 ......... <font color=blue>substitution</font>; use eq2
21 = q^2
q^2 = 21
q = <font color=red>sqrt(21)</font>
p - (1/p) = <font color=red>sqrt(21)</font>



------------------------------


Another approach would be to solve the equation p + (1/p) = 5 for p


p + (1/p) = 5
p * [ p + (1/p) ] = p*5
p^2 + 1 = 5p
p^2 - 5p + 1 = 0


Use the quadratic formula to find these two roots
p = (5 + sqrt(21))/2
p = (5 - sqrt(21))/2


Then use either root to compute p - (1/p)
I'll let you do these steps.


Hint:
If p = (5+sqrt(21))/2, then 1/p = (5 - sqrt(21))/2 after rationalizing the denominator.
A similar situation happens when p = (5 - sqrt(21))/2
</font>