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

Jun 19, 2018

$11 \pm 3 \sqrt{2}$

Explanation:

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

From the first equation:
$b = 22 - a$
Substitute into the second:
$a \left(22 - a\right) = 103$
$22 a - {a}^{2} = 103$
$0 = {a}^{2} - 22 a + 103$

$a = \frac{1}{2} \left(22 \pm \sqrt{484 - 412}\right) = \frac{1}{2} \left(22 \pm \sqrt{72}\right)$
$= \frac{1}{2} \left(22 \pm 6 \sqrt{2}\right) = 11 \pm 3 \sqrt{2}$

So we have two solutions for $a$. Use these to obtain $b$ from the first formula:
$11 \pm 3 \sqrt{2} + b = 22$
$b = 11 \overline{+} 3 \sqrt{2}$

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:
$11 \pm 3 \sqrt{2}$

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

Jun 19, 2018

$a = 11 + 3 \sqrt{2}$
$b = 11 - 3 \sqrt{2}$

Explanation:

$a + b = 22 \implies b = 22 - a$
$a b = 103$
$a \cdot \left(22 - a\right) = 103$
$- {a}^{2} + 22 a = 103$
${a}^{2} - 22 a + 103 = 0$
We have ${\left(a - 11\right)}^{2} - 121 + 103 = {\left(a - 11\right)}^{2} - 18$
$a - 11 = \pm \sqrt{18} = \pm 3 \sqrt{2}$
$a = 11 + 3 \sqrt{2}$
$b = 11 - 3 \sqrt{2}$