How do you find two positive real numbers that differ by 1 and have a product of 1?

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How do you find two positive real numbers that differ by 1 and have a product of 1?

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Answer 1 · 2016-03-01T19:23:48

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Explanation:

Assume $x$ and $y$ to be those positive real numbers such that $x<y$ .

Condition 1 : They differ by $1$ . That means $y-x=1$ ..... (1)
Condition 2: Their product is $1$ . That means $x.y=1$ ..... (2)

Use 1 ( $y=1+x$ ) to eliminate $y$ in 2
$x.(1+x)=1; \qquad => x^2+x-1=0$

Solve this quadratic equation
$x=\frac{-1\pm\sqrt{(1)^2-4(1)(-1)}}{2}$
$x_1= \frac{\sqrt{5}-1}{2}; x_2=-\frac{\sqrt{5}+1}{2}$

Of these two solutions we see that
$x_2 = -\frac{\sqrt{5}+1}{2}; \qquad y_2=1/x_2=-\frac{2}{\sqrt{5}+1}$ satisfies both conditions 1 and 2 and so is the solution.

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How do you find two positive real numbers that differ by 1 and have a product of 1?

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