Two positive numbers p and q have the property that their sum is equal to their product. If their difference is 7, what is \(\frac{1}{\frac{1}{p^2}+\frac{1}{q^2}}\)? Your answer will be of the form \(\frac{a+b\sqrt c}{d}\), where a and b don't both share the same common factor with d and c has no square as a factor. Find a+b+c+d.

qwerasdfzxcv1 Mar 4, 2020

#1**+1 **

We have that

pq = p + q (1)

p - q = 7 ⇒ p = 7 + q (2)

Sub (2) into (1) and we have that

(7 + q)q = 7 + 2q simplify

7q + q^2 = 7 + 2q rearrange

q^2 + 5q - 7 = 0 complete the square on q

q^2 + 5q + (25/4) = 7 + 25/4

(q + 5/2)^2 = 53/4 take the positive root

q + 5/2 = √53/2

q = √53/2 - 5/2 = [√53 - 5 ] / 2

q^2 = [ 53 - 10√53 + 25] / 4 = [ 78 - 10√53 ] / 4 = [39 - 5√53] / 2

So....p = √53/2 - 5/2 + 7 = √53/2 + 9/2 = [ √53 + 9 ] / 2

p^2 = [ 53 + 18√53 + 81] / 4 = [ 134 +18√53 ] / 4 = [ 67 + 9√53 ] / 2

Simplifying the given expression we have that

(pq)^2 (p + q)^2

_________ = ________

p^2 + q^2 p^2 + q^2

p + q = √53 + 2

(p+ q)^2 = ( 53 + 4√53 + 4) = (57 + 4√53)

p^2 + q^2 = [106 + 4√53 ] / 2 = 53 + 2√53

So we have

(57 + 4√53 ) (53 - 2√53) 2597 + 98√53

___________ ___________ = ______________

( 53 + 2√53) (53 - 2√53) 2597

So

a + b + c + d =

2597 + 98 + 53 + 2597 = 5345

CPhill Mar 5, 2020

#3**-1 **

Why don't you think it is right? (an explanation should have been given with your initial assertion)

Have you attempted to work through CPhill's logic?

Perhaps if there is an error you could be useful and tell him where it lay.

I see now you are a guest. Maybe not the original asker.

Melody
Mar 11, 2020