Hi Marj,

Suppose the two numbers are $x$ and $y.$ Their sum is 22 so

\[x + y = 22.\]

Their product is 125 and this gives you another equation. Solve the first equation for y, substitute into the second equation and simplify. Solve the resulting quadratic.

If you have a hard time believing the result use some computer software or a graphing calculator to plot the quadratic.

Penny

Eliminate one variable: $x+y = 22 \rightarrow y = 22-x$

Plug in: $125 = x(22-x);$

Tidy up: $x^2 - 22x + 125 = 0$

Now, the quadratic formula requires you to take the square root of $b^2 - 4ac$ where (here)$ a=1, b=-22, c = 125$

Can you do this? (If not, there is no solution.)

Good Hunting!
RD

Let x an y be the two real numbers with the desired properties
Then $x+y=22$ (sum is $22$) so $y = 22-x (*)$
And $xy=125$ (product is $125$) $(**)$
Substituting equation * into ** gives: $ x(22-x)=125$
Expanding gives: $22x - x^2 = 125$
Rearranging gives $x^2 - 22x + 125 = 0$ (if you knew the relationship between quadratic equations and their roots, then you could have jumped directly to this equation from the question)
Using the quadratic formula gives:

\[x = \frac{22 \pm \sqrt{484 - 500}}{2}\]
These generates two complex number roots.
Hence, there are no REAL numbers satisfying the desired properties!

Paul