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