Question

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

Answer 1

`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.

Answer 2

`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`