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)11±32

Explanation:

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

From the first equation:
b=22-ab=22a
Substitute into the second:
a(22-a)=103a(22a)=103
22a-a^2=10322aa2=103
0=a^2-22a+1030=a222a+103

Quadratic formula:
a=1/2(22+-sqrt(484-412))=1/2(22+-sqrt(72))a=12(22±484412)=12(22±72)
=1/2(22+-6sqrt(2))=11+-3sqrt(2)=12(22±62)=11±32

So we have two solutions for aa. Use these to obtain bb from the first formula:
11+-3sqrt(2)+b=2211±32+b=22
b=11bar(+)3sqrt(2)b=11¯¯¯¯¯+32

So for the pair of values of aa, bb always takes the other one of the two values. So the two numbers are simply the pair:
11+-3sqrt(2)11±32

Double check with the second formula that these are correct:
(11+3sqrt(2))(11-3sqrt(2))=103(11+32)(1132)=103
Note that this is a 'difference of two squares' formula
121-9*2=10312192=103
Yep, this checks out.

Jun 19, 2018

a=11+3sqrt2a=11+32
b=11-3sqrt2b=1132

Explanation:

a+b=22 => b=22-aa+b=22b=22a
ab=103ab=103
a*(22-a)=103a(22a)=103
-a^2+22a=103a2+22a=103
a^2-22a+103=0a222a+103=0
We have (a-11)^2-121+103=(a-11)^2-18(a11)2121+103=(a11)218
a-11=+-sqrt18=+-3sqrt2a11=±18=±32
a=11+3sqrt2a=11+32
b=11-3sqrt2b=1132