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Could anyone tell me why 1/b -b is taken as -1 since both the numbers differ from their reciprocals by 1 only

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vishumangal
Could anyone tell me why 1/b -b is taken as -1 since both the numbers differ from their reciprocals by 1 only

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The question says both the numbers differ from their reciprocals which means \(\frac{1}{a} - a = 1\) and \(\frac{1}{b} - b = 1\) but since both the numbers are different that's why one of the number will differ from its reciprocal by \(b - \frac{1}{b} = 1\) or we can rewrite as \(\frac{1}{b} - b = -1\).
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Bunuel
Two different positive numbers a and b each differ from their reciprocals by 1. What is a + b ?


(A) 1
(B) 2
(C) \(\sqrt{5}\)
(D) \(\sqrt{6}\)
(E) 3

It must be true that one number is more than its reciprocal, and the other is less than its reciprocal. We can let a be the one that is more than its reciprocal and hence b is the one that is less than its reciprocal. So we have:

a - 1/a = 1 and 1/b - b = 1

Subtracting these two equations, we have:

a - 1/a - 1/b + b = 0

a + b = 1/a + 1/b

a + b = (b + a)/(ab)

ab(a + b) = a + b

Since both a and b are positive, a + b ≠ 0. So we can divide both sides of the equation by a + b and obtain:

ab = 1

Now, let’s multiply the original first equation by a. We have:

a^2 - 1 = a

Similarly, let’s multiply the original second equation by b. We have:

1 - b^2 = b

Adding the two new equations, we have:

a^2 - b^2 = a + b

(a + b)(a - b) = a + b

Again, dividing both sides of the equation by a + b, we have:

a - b = 1

Squaring the above equation, we have:

a^2 - 2ab + b^2 = 1

Since ab = 1, we have:

a^2 - 2(1) + b^2 = 1

a^2 + b^2 = 3

Now add 2ab = 2 to the above equation, we have:

a^2 + 2ab + b^2 = 5

(a + b)^2 = 5

Taking the square root of both sides, we have:

a + b = √5

Answer: C
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Abhishek292kumar, hi can you please tell how did you come to the value of a as -1/2 + ((5)^(1/2))/2 ?
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abhishek31
hi can you please tell how did you come to the value of a as -1/2 + ((5)^(1/2))/2 ?
Hi,
By using the formula of quadratic equation:
\(a{x^2} + bx + c = 0\) then \(x =\frac { -b + \sqrt{{b^2}-4ac}}{2a}\) and \(x =\frac { -b - \sqrt{{b^2}-4ac}}{2a}\)
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Bunuel
Two different positive numbers a and b each differ from their reciprocals by 1. What is a + b ?


(A) 1
(B) 2
(C) \(\sqrt{5}\)
(D) \(\sqrt{6}\)
(E) 3

3/2 - 2/3 = 5/6
5/3 - 3/5 = 16/15

The red option yields a difference that is a little LESS than 1.
The blue option yields a difference that is a bit MORE than 1.
To yield a difference of 1, we need a combination of values between the red option and the blue option.
Since the blue difference is closer to 1 than is the red difference, the larger of the two values must be just a bit less than 5/3.

Implication:
One of the values ≈ 5/3 (since it differs from its reciprocal of 3/5 by about 1)
The other value ≈ 3/5 (since it differs from its reciprocal of 5/3 by about 1)

Thus:
a + b ≈ 5/3 + 3/5 = 34/15 ≈ 30/15 + 4/15 ≈ 2 + 1/4 = 2.25
The closest answer choice is C:
\(\sqrt{5} ≈ 2.24\)

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