Question #70499

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Answer 1 · 2017-10-20T18:12:02

#-3+4sqrt(2)# and #4sqrt(2)#

Let #x=#smaller number and #y=#larger number

So according to our first sentence:

#y-3=x->color(red)(y=x+3)#

Then our second sentence says:

#6y+x^2=41#

Then plug in #y# into our second equation:

#6(color(red)(x+3))+x^2=41#

Set this quadratic equation equal to #0#:

#6x+18+x^2=41#

#x^2+6x-23=0#

#x=(-6+-sqrt(6^2-4(1)(-23)))/(2(1))#

#x=(-6+-sqrt(36+92))/2=-6/2+-sqrt(128)/2=-3+-sqrt(64*2)/2#

#x=-3+-(8sqrt(2))/2=-3+-4sqrt(2)#

Since our two numbers are positive numbers, #-3-4sqrt(2)# cannot be our value for #x#, thus:

#x=color(blue)(-3+4sqrt(2))#

Plugging that in to our first equation:

#y=color(blue)(-3+4sqrt(2))+3=4sqrt(2)#

This number is also positive, so our two numbers are:

#-3+4sqrt(2)# and #4sqrt(2)#