# SOLUTION: instruction: find exact & approximate solutions to the problem. 106. One on One. Find two positive real numbers that differ by 1 and have a product of 1.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: instruction: find exact & approximate solutions to the problem. 106. One on One. Find two positive real numbers that differ by 1 and have a product of 1.      Log On

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 Quadratics: solvers Practice! Answers archive Lessons Word Problems In Depth

 Question 36378: instruction: find exact & approximate solutions to the problem. 106. One on One. Find two positive real numbers that differ by 1 and have a product of 1.Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!Find two positive real numbers..X AND Y SAY that differ by 1 and have a product of 1. X-Y=1..................................I XY=1 (X+Y)^2=(X-Y)^2+4XY=1^2+4*1=5 X+Y=SQRT(5).................................II EQN.I+EQN.II GIVES 2X=1+SQRT(5) X=(1+SQRT(5))/2=(2.236+1)/2=1.618 EQN.II-EQN.I..GIVES 2Y=SQRT(5)-1 Y=(SQRT(5)-1)/2=(2.236-1)/2=0.618