Question 72694
2 positive real numbers that differ by 1 and have a product of 1
:
Notice they didn't say that they were integers, just real, positive numbers.
:
Let one number = x; the other number = (x+1)
:
The product = 1
x(x+1) = 1
x^2 + x = 1
x^2 + x - 1 = 0; a quadratic equation
:
Use the quadratic formula: a = 1; b = 1; c = -1
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
:
{{{x = (-1 +- sqrt( 1^2-4*1*-1 ))/(2*1) }}}
:
{{{x = (-1 +- sqrt( 1- (-4)))/(2) }}}
:
{{{x = (-1 +- sqrt( 1 + 4 ))/(2) }}}
:
{{{x = (-1 - sqrt( 5 ))/(2) }}}
and
{{{x = (-1 + sqrt( 5 ))/(2) }}}; they only want the positive number for x
:
Check using decimals: {{{x = (-1 + sqrt( 5 ))/(2) }}} = .618 rounded
.618(1.618) = .9999, close enough