The following Property holds true for all real values of A and B
(a + b)^2 — (a - b)^2 = +4(a)(b)
If you perform the algebra and carry the minus sign through the 2nd expression, you will see that it is valid.
In this problem
Let X = a
Let (1/X) = b
We are given that: X + (1/X) = sqrt(40)
Or
(a + b) = sqrt(40)
Square both sides and you have:
(a + b)^2 = 40
Now put this into our template at the top
(40) — (a - b)^2 = +4(a)(b)
if we take A and multiply it by B:
(a) (b) = (X) (1/X) = (X / X) = 1
So substitute in 1 for (a)(b)
40 — (a - b)^2 = 4(1)
Rearrange the equation
36 = (a - b)^2
Take the square root of a variable expression squared— 6 will be equal to the Absolute Value of (a - b)
[a - b] = 6
And insert our terms
[X - (1/X)] = 6
Since X > 1 ——-> the quantity inside the Modulus will always be a positive quantity. x will be some value greater than 1 and (1/X) will be a positive proper fraction between 0 and 1
We can therefore remove the Modulus because it is redundant and we get
X + (1/X) = 6
Answer
6
C
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