+0

0
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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.

Mar 4, 2020

#1
+111435
+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

Mar 5, 2020
#2
-1

Cphil i dont think thats right

Guest Mar 5, 2020
#3
+110107
-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
edited by Melody  Mar 11, 2020
edited by Melody  Mar 11, 2020
#4
+110107
-2

Do not delete any of your questions.

It is completely unacceptable and could result in you being banned.

(This is in response to a question on a different post.)

Mar 11, 2020