Question 214334
Does anyone know how to go about this problem? The difference of two positive numbers is 3 and the sum of their squares is 65. Find the numbers. 


Step 1.  Let y be one positive and the larger number and x be the other positive and smaller number.


Step 2.  The problem statement provides the following relationships or equations:


{{{y-x=3}}} or {{{y=x+3}}}  (A)


{{{x^2+y^2=65}}}   (B)


Step 3.  Substitute (A) into (B)


{{{x^2+(x+3)^2=65}}}


{{{x^2+x^2+6x+9=65}}}


Step 4.  Subtract 65 from both sides of equation.


{{{x^2+x^2+6x+9-65=65-65}}}


{{{2x^2+6x-56=0}}}


{{{x^2+3x-28=0}}}


Step 5.  We can use the quadratic formula given as  {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


where a=1, b=3, and c=-28


*[invoke quadratic "x", 1,3, -28 ]


Step 6.  Based on the above steps, select 4 since we want a positive number.


With x=4, then y=3+4=7.


Check if {{{x^2+y^2=65}}}


{{{4^2+7^2=65}}}


{{{16+49=65}}}


{{{65=65}}} which is a true statement.


Step 7.  ANSWER:  The numbers are 4 and 7.


I hope the above steps were helpful. 


For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.


And good luck in your studies!


Respectfully,
Dr J