Two numbers have a sum of 22 and their product is 103. What are the numbers in simplest radical form?

2 Answers
Jun 19, 2018

#11+-3sqrt(2)#

Explanation:

Call the two numbers #a# and #b#. Then:
#a+b=22# and #ab=103#.

From the first equation:
#b=22-a#
Substitute into the second:
#a(22-a)=103#
#22a-a^2=103#
#0=a^2-22a+103#

Quadratic formula:
#a=1/2(22+-sqrt(484-412))=1/2(22+-sqrt(72))#
#=1/2(22+-6sqrt(2))=11+-3sqrt(2)#

So we have two solutions for #a#. Use these to obtain #b# from the first formula:
#11+-3sqrt(2)+b=22#
#b=11bar(+)3sqrt(2)#

So for the pair of values of #a#, #b# always takes the other one of the two values. So the two numbers are simply the pair:
#11+-3sqrt(2)#

Double check with the second formula that these are correct:
#(11+3sqrt(2))(11-3sqrt(2))=103#
Note that this is a 'difference of two squares' formula
#121-9*2=103#
Yep, this checks out.

Jun 19, 2018

#a=11+3sqrt2#
#b=11-3sqrt2#

Explanation:

#a+b=22 => b=22-a#
#ab=103#
#a*(22-a)=103#
#-a^2+22a=103#
#a^2-22a+103=0#
We have #(a-11)^2-121+103=(a-11)^2-18#
#a-11=+-sqrt18=+-3sqrt2#
#a=11+3sqrt2#
#b=11-3sqrt2#